Induction with two variables in PA

Question

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.

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 : $\forall x \ \forall y P(x,y)$. — Mauro ALLEGRANZA, Dec 02 '18 at 13:24
$P(n, 0)$ 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. — anon44508, Dec 02 '18 at 14:27
Nevermind, the other contributor cleared it all up. Thank you for your help. — anon44508, Dec 03 '18 at 02:36

score 0 · Accepted Answer · answered Dec 02 '18 at 20:55

0

Here's a proof in Fitch, which has a built in Peano Induction rule:

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

answered Dec 02 '18 at 20:55

Bram28

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Thanks for the mechanically verified proof! – anon44508 Dec 03 '18 at 02:33
@anon44508 You're welcome! :) – Bram28 Dec 03 '18 at 12:06

Induction with two variables in PA

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