2
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[EDIT/CONCLUSION] It turns out it was actually working.. I was just like too stupid to let the prover run for more time and assumed it would take a lot / not be able to prove with what I've provided because it wasn't doing it fast and because it took it ~1 minute with a similar goal asking for numbers lower or equal than 1. However, the goal I gave as an example in this post (leq(x,2)) took ~1 minute as well and for leq(x,5), it was actually able to prove it faster I think (~40 seconds if I recall correctly). Thank you very much to everyone who helped!!

[EDIT:] I apologize for not describing my axioms properly.. I have now written them so as they pretty much resemble what I actually wrote.

Hello and Happy New Year,

I am trying to define natural numbers in order to use them in a First-Order Logic Theorem Prover. The problem is I'm unable to solve the following problem:

"Show that the only numbers lower or equal than (for example) 2 are 0,1 and 2".

My axioms go like this: $$nat(0)$$ $$\forall x( nat(x) \to nat(s(x)) )$$ $$\forall x,y( (nat(x)\land nat(y)) \to nat(add(x,y))$$ $$\forall x( nat(x) \to (s(x) \neq 0) )$$ $$\forall x,y ( (nat(x) \land nat(y)) \to ( (s(x)=s(y)) \to (s=y)))$$ $$\forall x( (nat(x)\land x\neq 0) \to (\exists z (nat(z) \land (x = s(z)))))$$ $$\forall x(nat(x) \to (add(x,0) = x))$$ $$\forall x,y((nat(x)\land nat(y)) \to (add(x,s(y)) = s(add(x,y))))$$ $$\forall x,y((nat(x)\land nat(y)) \to ((leq(x,y) \leftrightarrow (\exists z(nat(z)\land (y = x + z) )))))$$

I have also added addition axioms for commutativity and associativity of addition.

So, what I'd like to be able to prove is something like this (for any number, not just for $2$):

$ \forall x(nat(x) \to (x\leq 2 \to (x = 0) \lor(x=1)\lor(x=2))) $.

Are additional axioms needed in order to be able to prove this?

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  • $\begingroup$ My answer was incorrect - I missed the axiom that every nonzero element has a successor. Editing . . . $\endgroup$ – Noah Schweber Jan 4 '17 at 0:09
  • $\begingroup$ I have edited the main post to more closely resemble the axioms I actually wrote in the program (it was nicer in the prover, because I could wrie something like this: $\forall \ nat(x) (P(x))$ instead of $\forall x (nat(x) \to P(x))$.. Regarding the linear order, no, the only axiom is that "definition" I wrote :). $\endgroup$ – PhantomR Jan 4 '17 at 0:23
  • $\begingroup$ I suggest, @Phantom that you define $nat(x)$, $s(x)$, and in a few of the axioms you provide, you need to use parentheses for anyone to understand what your axioms are trying to say. For example: $$\forall x,y (nat(x) \land nat(y) \to (s(x)=s(y) \to s=y)) $$ If $s(x)$ means the successor of $x$, than what does $s = y$ mean? Adding parentheses will further clarify exactly what you mean to say. $\endgroup$ – Namaste Jan 4 '17 at 19:11
  • $\begingroup$ Did you mean to write, in the axiom written above, to say that $$(s(x) = s(y)_ \rightarrow x = y\;$$ And if so, what I mean about using parentheses appropriately would be $$\forall x, y\Big(\big(nat(x) \land nat(y)\big) \rightarrow \big(((s(x) = s(y))\rightarrow (x=y))\big)\Big)$$ $\endgroup$ – Namaste Jan 4 '17 at 19:21
  • $\begingroup$ @amWhy Thank you for your advice! Indeed, I agree some of them are quite ambiguous. I will try to edit. $\endgroup$ – PhantomR Jan 4 '17 at 21:02
3
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These axioms are sufficient. Here's an outline of the proof in natural language:

Since $x\leq s(s(0))$, there exists $z$ such that $x + z = s(s(0))$.

Case 1: $z = 0$. Then $x = x + z = s(s(0))$, and we're done ($x = 2$). [Note that you've stated as an axiom $\text{add}(x,0) = 0$, and I assume you meant $\text{add}(x,0) = x$!]

Case 2: $z\neq 0$. Then there exists $y$ such that $z = s(y)$. Then $x + s(y) = s(s(0))$, so $s(x+y) = s(s(0))$, so $x+y = s(0)$.

We've managed to decrease the number of applications of $s$ by one. Now we repeat.

Case 1: $y = 0$. Then $x = x+y = s(0)$, and we're done ($x = 1$).

Case 2: $y \neq 0$. Then there exists $w$ such that $y = s(w)$. Then $x + s(w) = s(0)$, so $s(x+w) = s(0)$, so $x+w = 0$.

If $w = 0$, we're done, just as above ($x = 0$). Otherwise, $w = s(t)$, so $0 = x+ s(t) = s(x+t)$, and $0$ is a successor, contradiction.

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  • $\begingroup$ Thank you very much. I really liked the proof. However, the prover is seemingly unable to prove it (or it's taking a really, really long time..). Maybe the "nat" predicate is wrong? Should I've added some additional axioms for it? $\endgroup$ – PhantomR Jan 4 '17 at 18:53
  • $\begingroup$ Having zero experience with automated theorem provers, I can't give you any advice on how to speed it up. However, you should make sure that addition by $0$ axiom is correct, or your prover will never derive this sentence (and instead will derive a lot of false sentences)! $\endgroup$ – Alex Kruckman Jan 4 '17 at 18:56
  • $\begingroup$ Thank you for all your help. Sadly (I wish that were the problem), it was correctly written in the prover (I only mistook it here, thank you for pointing it out). I'll maybe try to read the axioms very carefully.. I may have written some of them wrongly. But basic arithmetic seemed to be working.. $\endgroup$ – PhantomR Jan 4 '17 at 19:07
  • $\begingroup$ @PhantomR I just completed a formal proof for this in the Fitch formal proof system and it checks out without any further axioms. The proof was 103 lines, including the 9 axioms, so 94 inference steps. I don't know what automated prover you are using, but maybe that's already too long to expect an automated prover to be able to find in a reasonable amount of time. $\endgroup$ – Bram28 Jan 4 '17 at 19:24
  • $\begingroup$ @Bram28 Thank you very much for your effort. If I could ask, does your prover have equality included (if yes, is that what you used)? Mine does, but maybe it would have been better to write my own equality predicate.. $\endgroup$ – PhantomR Jan 4 '17 at 19:50
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Here's a formal proof, copied and pasted from my Fitch software program (sorry for the lack of formatting!) which says it all checks out .. meaning that no: you don't need any further axioms (I did use all 9 axioms). It follows Alex's more informal proof:

  1. Nat(0) Axiom
  2. ∀x (Nat(x) → Nat(s(x))) Axiom
  3. ∀x ∀y ((Nat(x) ∧ Nat(y)) → Nat(x + y)) Axiom
  4. ∀x (Nat(x) → ¬s(x) = 0) Axiom
  5. ∀x ∀y ((Nat(x) ∧ Nat(y)) → (s(x) = s(y) → x = y)) Axiom
  6. ∀x ((Nat(x) ∧ ¬x = 0) → ∃z (Nat(z) ∧ x = s(z))) Axiom
  7. ∀x (Nat(x) → x + 0 = x) Axiom
  8. ∀x ∀y ((Nat(x) ∧ Nat(y)) → x + s(y) = s(x + y)) Axiom
  9. ∀x ∀y ((Nat(x) ∧ Nat(y)) → (Leq(x,y) ↔ ∃z (Nat(z) ∧ y = x + z))) Axiom
  10. Nat(0) → Nat(s(0)) ∀ Elim : 2
  11. Nat(s(0)) → Elim : 1, 10
  12. Nat(s(0)) → Nat(s(s(0))) ∀ Elim : 2
  13. Nat(s(s(0))) → Elim : 11, 12
  14. $\qquad$ a Nat(a) Assumption (introduces 'a' and assumes that 'a' is a Nat)
  15. $\qquad$ Nat(a) → a + 0 = a ∀ Elim : 7
  16. $\qquad$ a + 0 = a → Elim : 14, 15
  17. $\qquad \qquad$ Leq(a,s(s(0))) Assumption
  18. $\qquad \qquad$ (Nat(a) ∧ Nat(s(s(0)))) → (Leq(a,s(s(0))) ↔ ∃z (Nat(z) ∧ s(s(0)) = a + z)) ∀ Elim : 9
  19. $\qquad \qquad$ Nat(a) ∧ Nat(s(s(0))) ∧ Intro : 14, 13
  20. $\qquad \qquad$ Leq(a,s(s(0))) ↔ ∃z (Nat(z) ∧ s(s(0)) = a + z) → Elim : 18, 19
  21. $\qquad \qquad$ ∃z (Nat(z) ∧ s(s(0)) = a + z) ↔ Elim : 20, 17
  22. $\qquad \qquad \qquad$ b Nat(b) ∧ s(s(0)) = a + b Assumption (effectively: 'let 'b' be the 'z' from line 21')
  23. $\qquad \qquad \qquad$ Nat(b) ∧ Elim : 22
  24. $\qquad \qquad \qquad$ s(s(0)) = a + b ∧ Elim : 22
  25. $\qquad \qquad \qquad$ s(s(0)) = s(s(0)) = Intro
  26. $\qquad \qquad \qquad$ a + b = s(s(0)) = Elim : 24, 25
  27. $\qquad \qquad \qquad \qquad$ ¬(a = 0 ∨ a = s(0) ∨ a = s(s(0))) Assumption (setting up a Proof by Contradiction)
  28. $\qquad \qquad \qquad \qquad \qquad$ b = 0 Assumption
  29. $\qquad \qquad \qquad \qquad \qquad$ a + 0 = s(s(0)) = Elim : 28, 26
  30. $\qquad \qquad \qquad \qquad \qquad$ a = s(s(0)) = Elim : 16, 29
  31. $\qquad \qquad \qquad \qquad \qquad$ a = 0 ∨ a = s(0) ∨ a = s(s(0)) ∨ Intro : 30
  32. $\qquad \qquad \qquad \qquad \qquad$ ⊥ ⊥ Intro : 31, 27
  33. $\qquad \qquad \qquad \qquad$ ¬b = 0 ¬ Intro : 28-32
  34. $\qquad \qquad \qquad \qquad$ Nat(b) ∧ ¬b = 0 ∧ Intro : 23, 33
  35. $\qquad \qquad \qquad \qquad$ (Nat(b) ∧ ¬b = 0) → ∃z (Nat(z) ∧ b = s(z)) ∀ Elim : 6
  36. $\qquad \qquad \qquad \qquad$ ∃z (Nat(z) ∧ b = s(z)) → Elim : 34, 35
  37. $\qquad \qquad \qquad \qquad \qquad$ c Nat(c) ∧ b = s(c) Assumption (effectively: 'let 'c' be the 'z' from line 36')
  38. $\qquad \qquad \qquad \qquad \qquad$ Nat(c) ∧ Elim : 37
  39. $\qquad \qquad \qquad \qquad \qquad$ b = s(c) ∧ Elim : 37
  40. $\qquad \qquad \qquad \qquad \qquad$ a + s(c) = s(s(0)) = Elim : 39, 26
  41. $\qquad \qquad \qquad \qquad \qquad$ Nat(a) ∧ Nat(c) ∧ Intro : 14, 38
  42. $\qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(c)) → a + s(c) = s(a + c) ∀ Elim : 8
  43. $\qquad \qquad \qquad \qquad \qquad$ a + s(c) = s(a + c) → Elim : 41, 42
  44. $\qquad \qquad \qquad \qquad \qquad$ s(a + c) = s(s(0)) = Elim : 43, 40
  45. $\qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(c)) → Nat(a + c) ∀ Elim : 3
  46. $\qquad \qquad \qquad \qquad \qquad$ Nat(a + c) → Elim : 41, 45
  47. $\qquad \qquad \qquad \qquad \qquad$ Nat(a + c) ∧ Nat(s(0)) ∧ Intro : 46, 11
  48. $\qquad \qquad \qquad \qquad \qquad$ (Nat(a + c) ∧ Nat(s(0))) → (s(a + c) = s(s(0)) → a + c = s(0)) ∀ Elim : 5
  49. $\qquad \qquad \qquad \qquad \qquad$ s(a + c) = s(s(0)) → a + c = s(0) → Elim : 47, 48
  50. $\qquad \qquad \qquad \qquad \qquad$ a + c = s(0) → Elim : 44, 49
  51. $\qquad \qquad \qquad \qquad \qquad\qquad$ c = 0 Assumption
  52. $\qquad \qquad \qquad \qquad \qquad\qquad$ a + 0 = s(0) = Elim : 51, 50
  53. $\qquad \qquad \qquad \qquad \qquad\qquad$ a = s(0) = Elim : 16, 52
  54. $\qquad \qquad \qquad \qquad \qquad\qquad$ a = 0 ∨ a = s(0) ∨ a = s(s(0)) ∨ Intro : 53
  55. $\qquad \qquad \qquad \qquad \qquad\qquad$ ⊥ ⊥ Intro : 54, 27
  56. $\qquad \qquad \qquad \qquad \qquad$ ¬c = 0 ¬ Intro : 51-55
  57. $\qquad \qquad \qquad \qquad \qquad$ Nat(c) ∧ ¬c = 0 ∧ Intro : 38, 56
  58. $\qquad \qquad \qquad \qquad \qquad$ (Nat(c) ∧ ¬c = 0) → ∃z (Nat(z) ∧ c = s(z)) ∀ Elim : 6
  59. $\qquad \qquad \qquad \qquad \qquad$ ∃z (Nat(z) ∧ c = s(z)) → Elim : 57, 58
  60. $\qquad \qquad \qquad \qquad \qquad\qquad$ d Nat(d) ∧ c = s(d) Assumption (effectively: 'let 'd' be the 'z' from line 59')
  61. $\qquad \qquad \qquad \qquad \qquad\qquad$ Nat(d) ∧ Elim : 60
  62. $\qquad \qquad \qquad \qquad \qquad\qquad$ c = s(d) ∧ Elim : 60
  63. $\qquad \qquad \qquad \qquad \qquad \qquad$ a + s(d) = s(0) = Elim : 62, 50
  64. $\qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a) ∧ Nat(d) ∧ Intro : 14, 61
  65. $\qquad \qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(d)) → a + s(d) = s(a + d) ∀ Elim : 8
  66. $\qquad \qquad \qquad \qquad \qquad \qquad$ a + s(d) = s(a + d) → Elim : 64, 65
  67. $\qquad \qquad \qquad \qquad \qquad \qquad$ s(a + d) = s(0) = Elim : 66, 63
  68. $\qquad \qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(d)) → Nat(a + d) ∀ Elim : 3
  69. $\qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a + d) → Elim : 64, 68
  70. $\qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a + d) ∧ Nat(0) ∧ Intro : 69, 1
  71. $\qquad \qquad \qquad \qquad \qquad \qquad$ (Nat(a + d) ∧ Nat(0)) → (s(a + d) = s(0) → a + d = 0) ∀ Elim : 5
  72. $\qquad \qquad \qquad \qquad \qquad \qquad$ s(a + d) = s(0) → a + d = 0 → Elim : 70, 71
  73. $\qquad \qquad \qquad \qquad \qquad \qquad$ a + d = 0 → Elim : 67, 72
  74. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ d = 0 Assumption
  75. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ a + 0 = 0 = Elim : 74, 73
  76. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ a = 0 = Elim : 16, 75
  77. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ a = 0 ∨ a = s(0) ∨ a = s(s(0)) ∨ Intro : 76
  78. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ ⊥ ⊥ Intro : 77, 27
  79. $\qquad \qquad \qquad \qquad \qquad \qquad $ ¬d = 0 ¬ Intro : 74-78
  80. $\qquad \qquad \qquad \qquad \qquad \qquad $ Nat(d) ∧ ¬d = 0 ∧ Intro : 61, 79
  81. $\qquad \qquad \qquad \qquad \qquad \qquad $ (Nat(d) ∧ ¬d = 0) → ∃z (Nat(z) ∧ d = s(z)) ∀ Elim : 6
  82. $\qquad \qquad \qquad \qquad \qquad \qquad $ ∃z (Nat(z) ∧ d = s(z)) → Elim : 81, 80
  83. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ e Nat(e) ∧ d = s(e) Assumption (effectively: 'let 'e' be the 'z' from line 82')
  84. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ Nat(e) ∧ Elim : 83
  85. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ d = s(e) ∧ Elim : 83
  86. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ a + s(e) = 0 = Elim : 85, 73
  87. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a) ∧ Nat(e) ∧ Intro : 14, 84
  88. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(e)) → a + s(e) = s(a + e) ∀ Elim : 8
  89. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ a + s(e) = s(a + e) → Elim : 87, 88
  90. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ s(a + e) = 0 = Elim : 89, 86
  91. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ (Nat(a) ∧ Nat(e)) → Nat(a + e) ∀ Elim : 3
  92. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a + e) → Elim : 87, 91
  93. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ Nat(a + e) → ¬s(a + e) = 0 ∀ Elim : 4
  94. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ ¬s(a + e) = 0 → Elim : 92, 93
  95. $\qquad \qquad \qquad \qquad \qquad \qquad \qquad$ ⊥ ⊥ Intro : 90, 94
  96. $\qquad \qquad \qquad \qquad \qquad \qquad$ ⊥ ∃ Elim : 83-95, 82
  97. $\qquad \qquad \qquad \qquad \qquad$ ⊥ ∃ Elim : 60-96, 59
  98. $\qquad \qquad \qquad \qquad$ ⊥ ∃ Elim : 37-97, 36
  99. $\qquad \qquad \qquad$ ¬¬(a = 0 ∨ a = s(0) ∨ a = s(s(0))) ¬ Intro : 27-98
  100. $\qquad \qquad \qquad$ a = 0 ∨ a = s(0) ∨ a = s(s(0)) ¬ Elim : 99
  101. $\qquad \qquad$ a = 0 ∨ a = s(0) ∨ a = s(s(0)) ∃ Elim : 22-100, 21
  102. $\qquad$ Leq(a,s(s(0))) → (a = 0 ∨ a = s(0) ∨ a = s(s(0))) → Intro : 17-101
  103. ∀x (Nat(x) → (Leq(x,s(s(0))) → (x = 0 ∨ x = s(0) ∨ x = s(s(0))))) ∀ Intro : 14-102
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  • $\begingroup$ Wow.. thank you. My prover is actually able to prove that 0,1 are the only naturals lower or equal than 1... but it takes ~ a minute and says it derives like ~34 thousand clauses or so... it seems that the prover can be configured with regard to what proof techniques it uses.. I Think I'll try to ask a question about that in the automated theorem provers section. Thank you once more for your help! :) $\endgroup$ – PhantomR Jan 4 '17 at 20:35
  • $\begingroup$ @PhantomR Ah! So that confirms my suspicion that your original problem was just too complicated for an automated prover! :) $\endgroup$ – Bram28 Jan 4 '17 at 20:51
  • $\begingroup$ I think I was just too stupid to let it run for a bit more and assumed it would take a lot / not finish... It turns out it didn't take it that long and works.. it solves for leq(x,1) in ~1 minute, leq(x,2) in like the same amount of time.. ti was actually able to solve for leq(x,5) in less than that (~40 seconds I think). SO it actually worked ^_^. Thank you for your help! $\endgroup$ – PhantomR Jan 4 '17 at 20:58
  • $\begingroup$ @PhantomR Huh! It solves the 5 case faster than the 1 case?! That's really surprising. Well, glad it all works now. Thanks! $\endgroup$ – Bram28 Jan 4 '17 at 21:13

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