# （-a）×（-a）= a×a

121

1. $$a + （b + c ）= （a + b ）+ c一种+（b+C）=（一种+b）+Ca + (b + c) = (a + b) + c$$

2. $$a + 0 = a一种+0=一种a + 0 = a$$

3. $$a + （− a ）= 0一种+（-一种）=0a + (-a) = 0$$

4. $$a + b = b + a一种+b=b+一种a + b = b + a$$ *

5. $$a × （b × c ）= （a × b ）× c一种×（b×C）=（一种×b）×Ca \times (b \times c) = (a \times b) \times c$$

6. $$a × 1 = a一种×1个=一种a \times 1 = a$$

7. $$1 × a = a1个×一种=一种1 × a = a$$

8. $$a × （b + c ）= （a × b ）+ （a × c ）一种×（b+C）=（一种×b）+（一种×C）a \times (b + c) = (a \times b) + (a × c)$$

9. $$（b + c ）× a = （b × a ）+ （c × a ）（b+C）×一种=（b×一种）+（C×一种）(b + c) \times a = (b \times a) + (c \times a)$$

(a + b) + (c + d) = (a + (b + c)) + d Ax. 1


(a + b) + (c + d) = ((a + b) + c) + d Ax. 1
= (a + (b + c)) + d Ax. 1


Anthony用户已经提供了一个在线证明验证器 ，可以代替TIO。

## 样例证明

 -(-a) = (-(-a)) + 0          Ax. 2
= 0 + (-(-a))          Ax. 4
= (a + (-a)) + (-(-a)) Ax. 3
= a + ((-a) + (-(-a))) Ax. 1
= a + 0                Ax. 3
= a                    Ax. 2


## 计分

### 引理

Lemma:
a × 0 = 0

Proof (7 steps):
a × 0 = (a × 0) + 0                        Ax. 2 (1)
= (a × 0) + ((a × b) + (-(a × b)))   Ax. 3 (1)
= ((a × 0) + (a × b)) + (-(a × b))   Ax. 1 (1)
= (a × (0 + b)) + (-(a × b))         Ax. 8 (1)
= (a × (b + 0)) + (-(a × b))         Ax. 4 (1)
= (a × b) + (-(a × b))               Ax. 2 (1)
= 0                                  Ax. 3 (1)

Theorem:
(a × 0) + (b × 0) = 0

Proof (15 steps):
(a × 0) + (b × 0) = 0 + (b × 0)  Lemma (7)
= (b × 0) + 0  Ax. 4 (1)
= b × 0        Ax. 2 (1)
= 0            Lemma (7)


*：已经指出该公理不是严格证明该属性的必要条件，但是仍然可以使用它。

†：由于没有以期望的等式出现，因此使用这些公理的证明也不是最小的。那就是这些公理不能帮助证明期望的事实。仅出于完整性考虑，将它们包括在内。$$1个1个1$$

8

8
@Tahg您应该证明它并提交证明作为答案。这与您将在此处看到的大多数（如果不是全部）问题不同。
HyperNeutrino

8

Sparr

8

–totalhuman

34
@icrieverytim如果有帮助，可以将公理列表视为具有九个内置参数替换函数的编程语言，这是将特定输入转换为特定输出的函数的代码。
Sparr

47

## 18步

(-a)*(-a) = ((-a)*(-a))+0                                             Axiom 2
= ((-a)*(-a))+(((a*a)+(a*(-a)))+(-((a*a)+(a*(-a)))))        Axiom 3
= (((-a)*(-a))+((a*a)+(a*(-a))))+(-((a*a)+(a*(-a))))        Axiom 1
= (((a*a)+(a*(-a)))+((-a)*(-a)))+(-((a*a)+(a*(-a))))        Axiom 4
= ((a*a)+((a*(-a))+((-a)*(-a))))+(-((a*a)+(a*(-a))))        Axiom 1
= ((a*a)+((a+(-a))*(-a)))+(-((a*a)+(a*(-a))))               Axiom 9
= ((a*a)+(0*(-a)))+(-((a*a)+(a*(-a))))                      Axiom 3
= ((a*(a+0))+(0*(-a)))+(-((a*a)+(a*(-a))))                  Axiom 2
= ((a*(a+(a+(-a))))+(0*(-a)))+(-((a*a)+(a*(-a))))           Axiom 3
= (((a*a)+(a*(a+(-a))))+(0*(-a)))+(-((a*a)+(a*(-a))))       Axiom 8
= ((a*a)+((a*(a+(-a)))+(0*(-a))))+(-((a*a)+(a*(-a))))       Axiom 1
= (a*a)+(((a*(a+(-a)))+(0*(-a)))+(-((a*a)+(a*(-a)))))       Axiom 1
= (a*a)+((((a*a)+(a*(-a)))+(0*(-a)))+(-((a*a)+(a*(-a)))))   Axiom 8
= (a*a)+(((a*a)+((a*(-a))+(0*(-a))))+(-((a*a)+(a*(-a)))))   Axiom 1
= (a*a)+(((a*a)+((a+0)*(-a)))+(-((a*a)+(a*(-a)))))          Axiom 9
= (a*a)+(((a*a)+(a*(-a)))+(-((a*a)+(a*(-a)))))              Axiom 2
= (a*a)+0                                                   Axiom 3
= a*a                                                       Axiom 2


@Etoplay出于好奇，您是否在Prolog中编写了程序？
JalilCompaoré17年

23

2

SztupY

4
@SztupY Axiom 3 v + (-v) = 0v = ((a*a)+(a*(-a))您一步一步到达那里。
MT0

4

29

## 18步

a*a = a*a + 0                                                 A2
= a*a + ((a*(-a) + a*(-a)) + (-(a*(-a) + a*(-a))))        A3
= (a*a + (a*(-a) + a*(-a))) + (-(a*(-a) + a*(-a)))        A1
= (a*a + a*((-a) + (-a))) + (-(a*(-a) + a*(-a)))          A8
= a*(a + ((-a) + (-a))) + (-(a*(-a) + a*(-a)))            A8
= a*((a + (-a)) + (-a)) + (-(a*(-a) + a*(-a)))            A1
= a*(0 + (-a)) + (-(a*(-a) + a*(-a)))                     A3
= a*((-a) + 0) + (-(a*(-a) + a*(-a)))                     A4
= a*(-a) + (-(a*(-a) + a*(-a)))                           A2
= (a + 0)*(-a) + (-(a*(-a) + a*(-a)))                     A2
= (a + (a + (-a)))*(-a) + (-(a*(-a) + a*(-a)))            A3
= ((a + a) + (-a))*(-a) + (-(a*(-a) + a*(-a)))            A1
= ((-a) + (a + a))*(-a) + (-(a*(-a) + a*(-a)))            A4
= ((-a)*(-a) + (a + a)*(-a)) + (-(a*(-a) + a*(-a)))       A9
= ((-a)*(-a) + (a*(-a) + a*(-a))) + (-(a*(-a) + a*(-a)))  A9
= (-a)*(-a) + ((a*(-a) + a*(-a)) + (-(a*(-a) + a*(-a))))  A1
= (-a)*(-a) + 0                                           A3
= (-a)*(-a)                                               A2


2

28

# 29 26个步骤

(-a) × (-a) = ((-a) + 0) × (-a)                                                  Ax. 2
= ((-a) + (a + (-a))) × (-a)                                         Ax. 3
= ((a + (-a)) + (-a)) × (-a)                                         Ax. 4
= (a + ((-a) + (-a))) × (-a)                                         Ax. 1
= (a × (-a)) + (((-a) + (-a)) × (-a))                                Ax. 9
= (a × ((-a) + 0)) + (((-a) + (-a)) × (-a))                          Ax. 2
= (a × ((-a) + (a + (-a)))) + (((-a) + (-a)) × (-a))                 Ax. 3
= (a × ((a + (-a)) + (-a))) + (((-a) + (-a)) × (-a))                 Ax. 4
= (a × (a + ((-a) + (-a)))) + (((-a) + (-a)) × (-a))                 Ax. 1
= ((a × a) + (a × ((-a) + (-a)))) + (((-a) + (-a)) × (-a))           Ax. 8
= (a × a) + ((a × ((-a) + (-a))) + (((-a) + (-a)) × (-a)))           Ax. 1
= (a × a) + (((a × (-a)) + (a × (-a))) + (((-a) + (-a)) × (-a)))     Ax. 8
= (a × a) + (((a + a) × (-a)) + (((-a) + (-a)) × (-a)))              Ax. 9
= (a × a) + (((a + a) + ((-a) + (-a))) × (-a))                       Ax. 9
= (a × a) + ((((a + a) + (-a)) + (-a)) × (-a))                       Ax. 1
= (a × a) + (((a + (a + (-a))) + (-a)) × (-a))                       Ax. 1
= (a × a) + (((a + 0) + (-a)) × (-a))                                Ax. 3
= (a × a) + ((a + (-a)) × (-a))                                      Ax. 2
= (a × a) + (0 × (-a))                                               Ax. 3
= (a × a) + ((0 × (-a)) + 0)                                         Ax. 2
= (a × a) + ((0 × (-a)) + ((0 × (-a)) + (-(0 × (-a)))))              Ax. 3
= (a × a) + (((0 × (-a)) + (0 × (-a))) + (-(0 × (-a))))              Ax. 1
= (a × a) + (((0 + 0) × (-a)) + (-(0 × (-a))))                       Ax. 9
= (a × a) + ((0 × (-a)) + (-(0 × (-a))))                             Ax. 2
= (a × a) + 0                                                        Ax. 3
= (a × a)                                                            Ax. 2


2

14

# 18步

(-a)*(-a)
= (-a)*(-a) + 0                             [Axiom 2]
= (-a)*(-a) + ((-a)*a + -((-a)*a))          [Axiom 3]
= ((-a)*(-a) + (-a)*a) + -((-a)*a)          [Axiom 1]
= ((-a)*(-a) + ((-a) + 0)*a) + -((-a)*a)    [Axiom 2]
= ((-a)*(-a) + ((-a)*a + 0*a)) + -((-a)*a)  [Axiom 9]
= (((-a)*(-a) + (-a)*a) + 0*a) + -((-a)*a)  [Axiom 1]
= ((-a)*((-a) + a) + 0*a) + -((-a)*a)       [Axiom 8]
= ((-a)*(a + (-a)) + 0*a) + -((-a)*a)       [Axiom 4]
= ((-a)*0 + 0*a) + -((-a)*a)                [Axiom 3]
= (0*a + (-a)*0) + -((-a)*a)                [Axiom 4]
= ((a + (-a))*a + (-a)*0) + -((-a)*a)       [Axiom 3]
= ((a*a + (-a)*a) + (-a)*0) + -((-a)*a)     [Axiom 9]
= (a*a + ((-a)*a + (-a)*0)) + -((-a)*a)     [Axiom 1]
= (a*a + (-a)*(a + 0)) + -((-a)*a)          [Axiom 8]
= (a*a + (-a)*a) + -((-a)*a)                [Axiom 2]
= a*a + ((-a)*a + -((-a)*a))                [Axiom 1]
= a*a + 0                                   [Axiom 3]
= a*a                                       [Axiom 2]


9
A2: (-a) x (-a) = ((-a) + 0) x (-a)
A3:             = ((-a) + (a + (-a))) x (-a)
A9:             = ((-a) x (-a)) + ((a + (-a)) x (-a))
A4:             = ((-a) x (-a)) + (((-a) + a) x (-a))
A9:             = ((-a) x (-a)) + (((-a) x (-a)) + (a x (-a)))
A1:             = (((-a) x (-a)) + ((-a) x (-a))) + (a x (-a))
A2:             = (((-a) x (-a)) + ((-a) x (-a))) + (a x ((-a) + 0))
A3:             = (((-a) x (-a)) + ((-a) x (-a))) + (a x ((-a) + (a + (-a))))
A8:             = (((-a) x (-a)) + ((-a) x (-a))) + ((a x (-a)) + (a x (a + (-a))))
A8:             = (((-a) x (-a)) + ((-a) x (-a))) + ((a x (-a)) + ((a x a) + (a x (-a))))
A4:             = (((-a) x (-a)) + ((-a) x (-a))) + ((a x (-a)) + ((a x (-a)) + (a x a)))
A1:             = (((-a) x (-a)) + ((-a) x (-a))) + (((a x (-a)) + (a x (-a))) + (a x a))
A8:             = ((-a) x ((-a) + (-a))) + (((a x (-a)) + (a x (-a))) + (a x a))
A8:             = ((-a) x ((-a) + (-a))) + ((a x ((-a) + (-a))) + (a x a))
A1:             = (((-a) x ((-a) + (-a))) + (a x ((-a) + (-a)))) + (a x a)
A9:             = (((-a) + a) x ((-a) + (-a))) + (a x a)
A4:             = ((a + (-a)) x ((-a) + (-a))) + (a x a)
Lemma:          = (0 x ((-a) + (-a))) + (a x a)
A3:             = 0 + (a x a)
A4:             = (a x a) + 0
A2:             = (a x a)

Lemma: 0 = 0 x a

A3: 0 = (0 x a) + (-(0 x a))
A2:   = ((0 + 0) x a) + (-(0 x a))
A9:   = ((0 x a) + (0 x a)) + (-(0 x a))
A1:   = (0 x a) + ((0 x a) + (-(0 x a)))
A3:   = (0 x a) + 0
A2:   = (0 x a)


27 26个步骤 谢谢 Funky Computer Man注意到重复的行。

1

@FunkyComputerMan谢谢！你是对的; 我不确定写引理^^时的想法。并感谢您的编辑和评论。
JalilCompaoré17年

1
@JalilCompaoré我想您可以A3通过应用A2第二个（-a）而不是第一个开始来保存最后一个。不过，我不确定，因为我现在没有时间进行处理。
H.PWiz

7

# 6 + 7 + 7 + 6 + 3 = 29个步骤

Lemma 1. a*0=0 (6 steps)

0 = a*0 + -(a*0)  axiom 3
= a*(0+0) + -(a*0) axiom 2
= (a*0 + a*0) + -(a*0) axiom 8
= a*0 + (a*0 + -(a*0)) axiom 1
= a*0 + 0 axiom 3
= a*0 axiom 2

Lemma 2. a*(-b) = -(a*b) (7 steps)

a*(-b) = a*(-b) + 0 axiom 2
= a*(-b) + (a*b + -(a*b)) axiom 3
= (a*(-b) + a*b) + -(a*b) axiom 1
= a*(-b+b) + -(a*b) axiom 8
= a*0 + -(a*b) axiom 3
= 0 + -(a*b) lemma 1
= -(a*b) axiom 2

Lemma 3. (-a)*b = -(a*b) (7 steps)
same as above

Lemma 4. -(-(a)) = a (6 steps)

-(-a) = (-(-a)) + 0    axiom 2
= 0 + (-(-a))          axiom 4
= (a + (-a)) + (-(-a)) axiom 3
= a + ((-a) + (-(-a))) axiom 1
= a + 0                axiom 3
= a                    axiom 2

Theorem. -a*-a=0 (3 steps)

-a*-a = -(a*(-a)) lemma 3
= -(-(a*a)) lemma 2
= a*a lemma 4

Q.E.D.


3

HyperNeutrino

11
“定理-a * -a = 0”应为= a * a？
Sparr

2
@ H.PWiz我对使用引理的人没有问题，但是每次使用它们时，它们花费的时间就长了。我建议不要使用它们，因为它们会妨碍优化，但是就我而言，这篇文章很好。

4

Sparr

2

6

# 23个步骤

(-a) * (-a) = ((-a) * (-a)) + 0                                 ✔ axiom 2
= ((-a) * (-a)) + (((-a) * a) + -((-a) * a))        ✔ axiom 3
= (((-a) * (-a)) + (-a) * a) + -((-a) * a)          ✔ axiom 1
= (-a) * (-a + a) + -((-a) * a)                     ✔ axiom 8
= (-a) * (a + (-a)) + -((-a) * a)                   ✔ axiom 4
= ((-a) * 0) + -((-a) * a)                          ✔ axiom 3
= (((-a) * 0) + 0) + -((-a) * a)                    ✔ axiom 2
= ((-a) * 0 + ((-a)*0 + -((-a)*0))) + -((-a) * a)   ✔ axiom 3
= (((-a) * 0 + (-a)*0) + -((-a)*0)) + -((-a) * a)   ✔ axiom 1
= ((-a) * (0 + 0) + -((-a)*0)) + -((-a) * a)        ✔ axiom 8
= ((-a) * 0 + -((-a)*0)) + -((-a) * a)              ✔ axiom 2
= 0 + -((-a) * a)                                   ✔ axiom 3
= (0* a) + -(0*a) + -((-a) * a)                     ✔ axiom 3
= ((0+0)* a) + -(0*a) + -((-a) * a)                 ✔ axiom 2
= ((0 * a ) + (0*a) + -(0*a)) + -((-a) * a)         ✔ axiom 9
= ((0 * a ) + ((0*a) + -(0*a))) + -((-a) * a)       ✔ axiom 1
= ((0 * a ) + 0) + -((-a) * a)                      ✔ axiom 3
= (0 * a ) + -((-a) * a)                            ✔ axiom 2
= ((a + -a) * a ) + -((-a) * a)                     ✔ axiom 3
= ((a * a) + (-a) * a) + -((-a) * a)                ✔ axiom 9
= (a * a) + (((-a) * a) + -((-a) * a))              ✔ axiom 1
= (a * a) + 0                                       ✔ axiom 3
= a * a                                             ✔ axiom 2


5

# 34级

Lemma 1: 0=0*a (8 steps)
0
A3: a*0 + -(a*0)
A4: -(a*0) + a*0
A2: -(a*0) + a*(0+0)
A8: -(a*0) + (a*0 + a*0)
A1: (-(a*0) + a*0) + a*0
A3: 0 + a*0
A4: a*0 + 0
A2: a*0

Theorem: -a*-a = a*a (49 steps)

-a * -a
A2: (-a+0) * -a
A2: (-a+0) * (-a+0)
A3: (-a+(a+-a)) * (-a+0)
A3: (-a+(a+-a)) * (-a+(a+-a))
A8: -a*(-a+(a+-a)) + (a+-a)*(-a+(a+-a))
A8: -a*(-a+(a+-a)) + -a*(-a+(a+-a)) + a*(-a+(a+-a))
A3: -a*(-a+0)      + -a*(-a+(a+-a)) + a*(-a+(a+-a))
A3: -a*(-a+0)      + -a*(-a+0)      + a*(-a+(a+-a))
A8: -a*(-a+0)      + -a*(-a+0)      + a*-a + a*(a+-a)
A8: -a*(-a+0)      + -a*(-a+0)      + a*-a + a*a + a*-a
A2: -a*-a          + -a*(-a+0)      + a*-a + a*a + a*-a
A2: -a*-a          + -a*-a          + a*-a + a*a + a*-a
A8: -a*-a          + (-a+a)*-a             + a*a + a*-a
A3: -a*-a          + 0*-a                  + a*a + a*-a
L1: -a*-a          + 0                     + a*a + a*-a
A2: -a*-a                                  + a*a + a*-a
A4: a*a + -a*-a + a*-a
A8: a*a + (-a+a)*-a
A3: a*a + 0*-a
L1: a*a + 0
A2: a*a


1

Sparr

5

# 25步

0 = (0*a) + (-(0*a))       | Ax. 3
= ((0+0)*a) + (-(0*a))   | Ax. 2
= (0*a + 0*a) + (-(0*a)) | Ax. 9
= 0*a + (0*a + (-(0*a))) | Ax. 1
= 0*a + (0)              | Ax. 3
= 0*a                    | Ax. 2


Lemma M [12个步骤](-a)*b = -(a*b)

(-a)*b = (-a)*b + 0                | Ax. 2
= (-a)*b + (a*b + (-(a*b))) | Ax. 3
= ((-a)*b + a*b) + (-(a*b)) | Ax. 5
= ((-a)+a)*b + (-(a*b))     | Ax. 9
= 0*b + (-(a*b))            | Ax. 3
= 0 + (-(a*b))              | Lem. Z 
= -(a*b)                    | Ax. 2


(-a)*(-a) = (-a)*(-a) + 0                | Ax. 2
= 0 + (-a)*(-a)                | Ax. 4
= (a*a + (-(a*a))) + (-a)*(-a) | Ax. 3
= a*a + ((-(a*a)) + (-a)*(-a)) | Ax. 1
= a*a + ((-a)*a + (-a)*(-a))   | Lem. M 
= a*a + ((-a)*(a + (-a)))      | Ax. 8
= a*a + ((-a)*0)               | Ax. 3
= a*a + 0                      | Lem. Z 
= a*a                          | Ax. 2


SmileAndNod

4

## 22 23个步骤

• 这个问题不允许在方程的两边都加上项。相反，我们只能修改初始字符串。
• 乘法不被认为是可交换的。
• 给我们一个单元1，但是它在拼图中没有任何作用，因为它只涉及定义它的规则。

(-a)×(-a) :=
n×n =
n×n + 0 =                                [Ax. 2]
n×n + [n×a + -(n×a)] =                   [Ax. 3]
[n×n + n×a] + -(n×a) =                   [Ax. 1]
[n×(n+a)] + -(n×a) =                     [Ax. 8]
[n×(n+a) + 0] + -(n×a) =                 [Ax. 2]
[n×(n+a) + (n×a + -(n×a))] + -(n×a) =    [Ax. 3]
[(n×(n+a) + n×a) + -(n×a)] + -(n×a) =    [Ax. 1]
[n×((n+a) + a) + -(n×a)] + -(n×a) =      [Ax. 8]
[n×((a+n) + a) + -(n×a)] + -(n×a) =      [Ax. 4]
[n×(0 + a) + -(n×a)] + -(n×a) =          [Ax. 3]
[n×(a + 0) + -(n×a)] + -(n×a) =          [Ax. 4]
[n×a + -(n×a)] + -(n×a) =                [Ax. 2]
[(n+0)×a + -(n×a)] + -(n×a) =            [Ax. 2]
[(0+n)×a + -(n×a)] + -(n×a) =            [Ax. 4]
[((a+n)+n)×a + -(n×a)] + -(n×a) =        [Ax. 3]
[((a+n)×a+n×a) + -(n×a)] + -(n×a) =      [Ax. 9]
[(a+n)×a+(n×a + -(n×a))] + -(n×a) =      [Ax. 1]
[(a+n)×a + 0] + -(n×a) =                 [Ax. 3]
[(a+n)×a] + -(n×a) =                     [Ax. 2]
[a×a+n×a] + -(n×a) =                     [Ax. 9]
a×a+[n×a + -(n×a)] =                     [Ax. 1]
a×a+0 =                                  [Ax. 3]
a×a                                      [Ax. 2]


@ H.PWiz你为什么不能从去n0 + n一步到位？不只是A2吗？规则确实说过，变量也可以代表任意复杂的表达式
jq170727 '17

@ jq170727 Axiom 2仅指出了这a + 0 = a一点0 + a = a。你需要一个额外的交换步骤就可以得到n0 + n

@ H.PWiz您不能反向阅读公理吗？
jq170727

1
@ jq170727不，您不必为此使用可交换性。
JalilCompaoré17年

4

# 304步

ringAxioms = {ForAll[{a, b, c}, p[a, p[b, c]] == p[p[a, b], c]],
ForAll[a, p[a, 0] == a],
ForAll[a, p[a, n[a]] == 0],
ForAll[{a, b}, p[a, b] == p[b, a]],
ForAll[{a, b, c}, t[a, t[b, c]] == t[t[a, b], c]],
ForAll[a, t[a, 1] == a], ForAll[a, t[1, a] == a],
ForAll[{a, b, c}, t[a, p[b, c]] == p[t[a, b], t[a, c]]],
ForAll[{a, b, c}, t[p[b, c], a] == p[t[b, a], t[c, a]]]};

proof = FindEquationalProof[t[n[a], n[a]] == t[a, a], ringAxioms];

proof["ProofNotebook"]


graph = proof["ProofGraph"];
score = Sum[
Length[FindPath[graph, axiom, "Conclusion 1", Infinity,
All]], {axiom,
Select[VertexList[graph], StringMatchQ["Axiom " ~~ __]]}]


Axiom 1

We are given that:

x1==p[x1, 0]

Axiom 2

We are given that:

x1==t[x1, 1]

Axiom 3

We are given that:

x1==t[1, x1]

Axiom 4

We are given that:

p[x1, x2]==p[x2, x1]

Axiom 5

We are given that:

p[x1, p[x2, x3]]==p[p[x1, x2], x3]

Axiom 6

We are given that:

p[x1, n[x1]]==0

Axiom 7

We are given that:

p[t[x1, x2], t[x3, x2]]==t[p[x1, x3], x2]

Axiom 8

We are given that:

p[t[x1, x2], t[x1, x3]]==t[x1, p[x2, x3]]

Axiom 9

We are given that:

t[x1, t[x2, x3]]==t[t[x1, x2], x3]

Hypothesis 1

We would like to show that:

t[n[a], n[a]]==t[a, a]

Critical Pair Lemma 1

The following expressions are equivalent:

p[0, x1]==x1

Proof

Note that the input for the rule:

p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, 0]->x1

where these rules follow from Axiom 4 and Axiom 1 respectively.

Critical Pair Lemma 2

The following expressions are equivalent:

p[x1, p[n[x1], x2]]==p[0, x2]

Proof

Note that the input for the rule:

p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, n[x1_]]->0

where these rules follow from Axiom 5 and Axiom 6 respectively.

Critical Pair Lemma 3

The following expressions are equivalent:

t[p[1, x1], x2]==p[x2, t[x1, x2]]

Proof

Note that the input for the rule:

p[t[x1_, x2_], t[x3_, x2_]]->t[p[x1, x3], x2]

contains a subpattern of the form:

t[x1_, x2_]

which can be unified with the input for the rule:

t[1, x1_]->x1

where these rules follow from Axiom 7 and Axiom 3 respectively.

Critical Pair Lemma 4

The following expressions are equivalent:

t[x1, p[1, x2]]==p[x1, t[x1, x2]]

Proof

Note that the input for the rule:

p[t[x1_, x2_], t[x1_, x3_]]->t[x1, p[x2, x3]]

contains a subpattern of the form:

t[x1_, x2_]

which can be unified with the input for the rule:

t[x1_, 1]->x1

where these rules follow from Axiom 8 and Axiom 2 respectively.

Critical Pair Lemma 5

The following expressions are equivalent:

t[p[1, x1], 0]==t[x1, 0]

Proof

Note that the input for the rule:

p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]

contains a subpattern of the form:

p[x1_, t[x2_, x1_]]

which can be unified with the input for the rule:

p[0, x1_]->x1

where these rules follow from Critical Pair Lemma 3 and Critical Pair Lemma 1 respectively.

Critical Pair Lemma 6

The following expressions are equivalent:

t[0, 0]==t[1, 0]

Proof

Note that the input for the rule:

t[p[1, x1_], 0]->t[x1, 0]

contains a subpattern of the form:

p[1, x1_]

which can be unified with the input for the rule:

p[x1_, 0]->x1

where these rules follow from Critical Pair Lemma 5 and Axiom 1 respectively.

Substitution Lemma 1

It can be shown that:

t[0, 0]==0

Proof

We start by taking Critical Pair Lemma 6, and apply the substitution:

t[1, x1_]->x1

which follows from Axiom 3.

Critical Pair Lemma 7

The following expressions are equivalent:

t[x1, 0]==t[p[x1, 1], 0]

Proof

Note that the input for the rule:

t[p[1, x1_], 0]->t[x1, 0]

contains a subpattern of the form:

p[1, x1_]

which can be unified with the input for the rule:

p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]

where these rules follow from Critical Pair Lemma 5 and Axiom 4 respectively.

Critical Pair Lemma 8

The following expressions are equivalent:

t[0, p[1, x1]]==t[0, x1]

Proof

Note that the input for the rule:

p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]

contains a subpattern of the form:

p[x1_, t[x1_, x2_]]

which can be unified with the input for the rule:

p[0, x1_]->x1

where these rules follow from Critical Pair Lemma 4 and Critical Pair Lemma 1 respectively.

Critical Pair Lemma 9

The following expressions are equivalent:

t[p[x1, 1], p[1, 0]]==p[p[x1, 1], t[x1, 0]]

Proof

Note that the input for the rule:

p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]

contains a subpattern of the form:

t[x1_, x2_]

which can be unified with the input for the rule:

t[p[x1_, 1], 0]->t[x1, 0]

where these rules follow from Critical Pair Lemma 4 and Critical Pair Lemma 7 respectively.

Substitution Lemma 2

It can be shown that:

t[p[x1, 1], 1]==p[p[x1, 1], t[x1, 0]]

Proof

We start by taking Critical Pair Lemma 9, and apply the substitution:

p[x1_, 0]->x1

which follows from Axiom 1.

Substitution Lemma 3

It can be shown that:

p[x1, 1]==p[p[x1, 1], t[x1, 0]]

Proof

We start by taking Substitution Lemma 2, and apply the substitution:

t[x1_, 1]->x1

which follows from Axiom 2.

Substitution Lemma 4

It can be shown that:

p[x1, 1]==p[x1, p[1, t[x1, 0]]]

Proof

We start by taking Substitution Lemma 3, and apply the substitution:

p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]

which follows from Axiom 5.

Critical Pair Lemma 10

The following expressions are equivalent:

t[0, x1]==t[0, p[x1, 1]]

Proof

Note that the input for the rule:

t[0, p[1, x1_]]->t[0, x1]

contains a subpattern of the form:

p[1, x1_]

which can be unified with the input for the rule:

p[x1_, x2_]\[TwoWayRule]p[x2_, x1_]

where these rules follow from Critical Pair Lemma 8 and Axiom 4 respectively.

Critical Pair Lemma 11

The following expressions are equivalent:

t[p[1, 0], p[x1, 1]]==p[p[x1, 1], t[0, x1]]

Proof

Note that the input for the rule:

p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]

contains a subpattern of the form:

t[x2_, x1_]

which can be unified with the input for the rule:

t[0, p[x1_, 1]]->t[0, x1]

where these rules follow from Critical Pair Lemma 3 and Critical Pair Lemma 10 respectively.

Substitution Lemma 5

It can be shown that:

t[1, p[x1, 1]]==p[p[x1, 1], t[0, x1]]

Proof

We start by taking Critical Pair Lemma 11, and apply the substitution:

p[x1_, 0]->x1

which follows from Axiom 1.

Substitution Lemma 6

It can be shown that:

p[x1, 1]==p[p[x1, 1], t[0, x1]]

Proof

We start by taking Substitution Lemma 5, and apply the substitution:

t[1, x1_]->x1

which follows from Axiom 3.

Substitution Lemma 7

It can be shown that:

p[x1, 1]==p[x1, p[1, t[0, x1]]]

Proof

We start by taking Substitution Lemma 6, and apply the substitution:

p[p[x1_, x2_], x3_]->p[x1, p[x2, x3]]

which follows from Axiom 5.

Substitution Lemma 8

It can be shown that:

p[x1, p[n[x1], x2]]==x2

Proof

We start by taking Critical Pair Lemma 2, and apply the substitution:

p[0, x1_]->x1

which follows from Critical Pair Lemma 1.

Critical Pair Lemma 12

The following expressions are equivalent:

n[n[x1]]==p[x1, 0]

Proof

Note that the input for the rule:

p[x1_, p[n[x1_], x2_]]->x2

contains a subpattern of the form:

p[n[x1_], x2_]

which can be unified with the input for the rule:

p[x1_, n[x1_]]->0

where these rules follow from Substitution Lemma 8 and Axiom 6 respectively.

Substitution Lemma 9

It can be shown that:

n[n[x1]]==x1

Proof

We start by taking Critical Pair Lemma 12, and apply the substitution:

p[x1_, 0]->x1

which follows from Axiom 1.

Critical Pair Lemma 13

The following expressions are equivalent:

x1==p[n[x2], p[x2, x1]]

Proof

Note that the input for the rule:

p[x1_, p[n[x1_], x2_]]->x2

contains a subpattern of the form:

n[x1_]

which can be unified with the input for the rule:

n[n[x1_]]->x1

where these rules follow from Substitution Lemma 8 and Substitution Lemma 9 respectively.

Critical Pair Lemma 14

The following expressions are equivalent:

t[x1, x2]==p[n[x2], t[p[1, x1], x2]]

Proof

Note that the input for the rule:

p[n[x1_], p[x1_, x2_]]->x2

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, t[x2_, x1_]]->t[p[1, x2], x1]

where these rules follow from Critical Pair Lemma 13 and Critical Pair Lemma 3 respectively.

Critical Pair Lemma 15

The following expressions are equivalent:

t[x1, x2]==p[n[x1], t[x1, p[1, x2]]]

Proof

Note that the input for the rule:

p[n[x1_], p[x1_, x2_]]->x2

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, t[x1_, x2_]]->t[x1, p[1, x2]]

where these rules follow from Critical Pair Lemma 13 and Critical Pair Lemma 4 respectively.

Critical Pair Lemma 16

The following expressions are equivalent:

p[1, t[x1, 0]]==p[n[x1], p[x1, 1]]

Proof

Note that the input for the rule:

p[n[x1_], p[x1_, x2_]]->x2

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, p[1, t[x1_, 0]]]->p[x1, 1]

where these rules follow from Critical Pair Lemma 13 and Substitution Lemma 4 respectively.

Substitution Lemma 10

It can be shown that:

p[1, t[x1, 0]]==1

Proof

We start by taking Critical Pair Lemma 16, and apply the substitution:

p[n[x1_], p[x1_, x2_]]->x2

which follows from Critical Pair Lemma 13.

Critical Pair Lemma 17

The following expressions are equivalent:

t[t[x1, 0], 0]==t[1, 0]

Proof

Note that the input for the rule:

t[p[1, x1_], 0]->t[x1, 0]

contains a subpattern of the form:

p[1, x1_]

which can be unified with the input for the rule:

p[1, t[x1_, 0]]->1

where these rules follow from Critical Pair Lemma 5 and Substitution Lemma 10 respectively.

Substitution Lemma 11

It can be shown that:

t[x1, t[0, 0]]==t[1, 0]

Proof

We start by taking Critical Pair Lemma 17, and apply the substitution:

t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]

which follows from Axiom 9.

Substitution Lemma 12

It can be shown that:

t[x1, 0]==t[1, 0]

Proof

We start by taking Substitution Lemma 11, and apply the substitution:

t[0, 0]->0

which follows from Substitution Lemma 1.

Substitution Lemma 13

It can be shown that:

t[x1, 0]==0

Proof

We start by taking Substitution Lemma 12, and apply the substitution:

t[1, x1_]->x1

which follows from Axiom 3.

Critical Pair Lemma 18

The following expressions are equivalent:

t[x1, t[0, x2]]==t[0, x2]

Proof

Note that the input for the rule:

t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]

contains a subpattern of the form:

t[x1_, x2_]

which can be unified with the input for the rule:

t[x1_, 0]->0

where these rules follow from Axiom 9 and Substitution Lemma 13 respectively.

Critical Pair Lemma 19

The following expressions are equivalent:

p[1, t[0, x1]]==p[n[x1], p[x1, 1]]

Proof

Note that the input for the rule:

p[n[x1_], p[x1_, x2_]]->x2

contains a subpattern of the form:

p[x1_, x2_]

which can be unified with the input for the rule:

p[x1_, p[1, t[0, x1_]]]->p[x1, 1]

where these rules follow from Critical Pair Lemma 13 and Substitution Lemma 7 respectively.

Substitution Lemma 14

It can be shown that:

p[1, t[0, x1]]==1

Proof

We start by taking Critical Pair Lemma 19, and apply the substitution:

p[n[x1_], p[x1_, x2_]]->x2

which follows from Critical Pair Lemma 13.

Critical Pair Lemma 20

The following expressions are equivalent:

t[0, t[0, x1]]==t[0, 1]

Proof

Note that the input for the rule:

t[0, p[1, x1_]]->t[0, x1]

contains a subpattern of the form:

p[1, x1_]

which can be unified with the input for the rule:

p[1, t[0, x1_]]->1

where these rules follow from Critical Pair Lemma 8 and Substitution Lemma 14 respectively.

Substitution Lemma 15

It can be shown that:

t[0, x1]==t[0, 1]

Proof

We start by taking Critical Pair Lemma 20, and apply the substitution:

t[x1_, t[0, x2_]]->t[0, x2]

which follows from Critical Pair Lemma 18.

Substitution Lemma 16

It can be shown that:

t[0, x1]==0

Proof

We start by taking Substitution Lemma 15, and apply the substitution:

t[x1_, 1]->x1

which follows from Axiom 2.

Critical Pair Lemma 21

The following expressions are equivalent:

t[n, x1]==p[n[x1], t[0, x1]]

Proof

Note that the input for the rule:

p[n[x1_], t[p[1, x2_], x1_]]->t[x2, x1]

contains a subpattern of the form:

p[1, x2_]

which can be unified with the input for the rule:

p[x1_, n[x1_]]->0

where these rules follow from Critical Pair Lemma 14 and Axiom 6 respectively.

Substitution Lemma 17

It can be shown that:

t[n, x1]==p[n[x1], 0]

Proof

We start by taking Critical Pair Lemma 21, and apply the substitution:

t[0, x1_]->0

which follows from Substitution Lemma 16.

Substitution Lemma 18

It can be shown that:

t[n, x1]==n[x1]

Proof

We start by taking Substitution Lemma 17, and apply the substitution:

p[x1_, 0]->x1

which follows from Axiom 1.

Critical Pair Lemma 22

The following expressions are equivalent:

t[n, t[x1, x2]]==t[n[x1], x2]

Proof

Note that the input for the rule:

t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]

contains a subpattern of the form:

t[x1_, x2_]

which can be unified with the input for the rule:

t[n, x1_]->n[x1]

where these rules follow from Axiom 9 and Substitution Lemma 18 respectively.

Substitution Lemma 19

It can be shown that:

n[t[x1, x2]]==t[n[x1], x2]

Proof

We start by taking Critical Pair Lemma 22, and apply the substitution:

t[n, x1_]->n[x1]

which follows from Substitution Lemma 18.

Critical Pair Lemma 23

The following expressions are equivalent:

t[x1, n]==p[n[x1], t[x1, 0]]

Proof

Note that the input for the rule:

p[n[x1_], t[x1_, p[1, x2_]]]->t[x1, x2]

contains a subpattern of the form:

p[1, x2_]

which can be unified with the input for the rule:

p[x1_, n[x1_]]->0

where these rules follow from Critical Pair Lemma 15 and Axiom 6 respectively.

Substitution Lemma 20

It can be shown that:

t[x1, n]==p[n[x1], 0]

Proof

We start by taking Critical Pair Lemma 23, and apply the substitution:

t[x1_, 0]->0

which follows from Substitution Lemma 13.

Substitution Lemma 21

It can be shown that:

t[x1, n]==n[x1]

Proof

We start by taking Substitution Lemma 20, and apply the substitution:

p[x1_, 0]->x1

which follows from Axiom 1.

Critical Pair Lemma 24

The following expressions are equivalent:

n[t[x1, x2]]==t[x1, t[x2, n]]

Proof

Note that the input for the rule:

t[x1_, n]->n[x1]

contains a subpattern of the form:

t[x1_, n]

which can be unified with the input for the rule:

t[t[x1_, x2_], x3_]->t[x1, t[x2, x3]]

where these rules follow from Substitution Lemma 21 and Axiom 9 respectively.

Substitution Lemma 22

It can be shown that:

t[n[x1], x2]==t[x1, t[x2, n]]

Proof

We start by taking Critical Pair Lemma 24, and apply the substitution:

n[t[x1_, x2_]]->t[n[x1], x2]

which follows from Substitution Lemma 19.

Substitution Lemma 23

It can be shown that:

t[n[x1], x2]==t[x1, n[x2]]

Proof

We start by taking Substitution Lemma 22, and apply the substitution:

t[x1_, n]->n[x1]

which follows from Substitution Lemma 21.

Substitution Lemma 24

It can be shown that:

t[a, n[n[a]]]==t[a, a]

Proof

We start by taking Hypothesis 1, and apply the substitution:

t[n[x1_], x2_]->t[x1, n[x2]]

which follows from Substitution Lemma 23.

Conclusion 1

We obtain the conclusion:

True

Proof

Take Substitution Lemma 24, and apply the substitution:

n[n[x1_]]->x1

which follows from Substitution Lemma 9.