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This probably has been asked before, but apologies, I don't know how to locate it. I want to prove $\forall x,y: P(x, y)$. My premises are:

$$P(0, 0) \wedge \\ [\forall x: P(x, 0)] \wedge \\ [\forall y: P(0, y)] \wedge \\ [\forall x,y: P(x, y) \Rightarrow P(Suc(x), Suc(y))]$$

I can prove some basic facts in PA, such as addition is comm, assoc, etc. I can also prove things about less than, such as less than is transitive. Also, if x is less than or equal to y, then there exists a z such x + z = y. But I'm not good at induction with two variables, and so cannot complete the proof.

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  • $\begingroup$ Not necessarily by double induction. You can prove $\forall y P(n,y)$ by induction on $y$, with $n$ whatever. If the proof works without any specific assumption regarding $n$, you can generalize on it, getting : $\forall x \ \forall y P(x,y)$. $\endgroup$ – Mauro ALLEGRANZA Dec 2 '18 at 13:24
  • $\begingroup$ See also the post : Induction on two integer variables. $\endgroup$ – Mauro ALLEGRANZA Dec 2 '18 at 13:27
  • $\begingroup$ See also : Mathematical induction: variants and subtleties. $\endgroup$ – Mauro ALLEGRANZA Dec 2 '18 at 13:28
  • $\begingroup$ $P(n, 0)$ is proven by the second assumption. However, I don't see how $\forall y : P(n, y) \Rightarrow P(n, Suc(y))$ is proven, because I only have $\forall n, y : P(n, y) \Rightarrow P(Suc(n), Suc(y))$. What am I not grasping? Also, I've seen those two links before, but thank you for sending them as they definitely contain helpful tricks for proof by induction. $\endgroup$ – anon44508 Dec 2 '18 at 14:27
  • $\begingroup$ Nevermind, the other contributor cleared it all up. Thank you for your help. $\endgroup$ – anon44508 Dec 3 '18 at 2:36
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Here's a proof in Fitch, which has a built in Peano Induction rule:

enter image description here

As you can see, the only 'trick' is to just ignore the inductive hypothesis for the inside inductive proof, but instead to use the inductive hypothesis for the outside inductive proof

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  • $\begingroup$ Thanks for the mechanically verified proof! $\endgroup$ – anon44508 Dec 3 '18 at 2:33
  • $\begingroup$ @anon44508 You're welcome! :) $\endgroup$ – Bram28 Dec 3 '18 at 12:06

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