Is induction like this logically rigorous?

Question

Suppose I want to prove $p(m,n)=1/2$ for all $m\ge1,n\ge1$, and I know the base case $p(1,1)=1/2$ holds.

Here is my induction proof:

1. assume $p(m-1,n),p(m,n-1),p(m-1,n-1),p(m-2,n-1)\cdots$ all equals to $1/2$.
2. with above assumption, show that $p(m,n)=1/2$.

Is above induction logically rigorous? why and why not?

Edit:

This question is my analogy to a single variable problem, which is like this:

Suppose we want to prove $p(n)=1/2$ holds for any positive integer $n$, it is easy to show that $p(1)=1/2$. Assuming $p(k)=1/2$ holds for all $k<n$, using this, prove $p(n)=1/2$.

This is the standard way to make an induction proof.

Answer 1

Really it depends on what is happening along the edges (when $m$ or $n$ is $1$). If your induction argument requires for instance that $p(2,0)$ be equal to $\frac12$ in order for $p(2,1)$ to be equal to $\frac12$, then your argument is not valid.

For instance $p(1,1) = \frac12$ and $p(m,n) = 5$ for all $(m,n)\ne (1,1)$ would be a counterexample.

Answer 2

The following induction principle is sound:

Suppose that $p(1,1)$ holds, and, for all $u,v \in \mathbb{N}$, if $p(u,v)$ holds then $p(u+1,v)$ and $p(u,v+1)$ hold. Then $p(n,m)$ holds for all $n,m \in \mathbb{N}$.

To see that the principle is sound, notice that for any $n,m \in \mathbb{N}$ we have a chain $$ p(1,1), p(2,1), \ldots, p(m,1), p(m,2), \ldots, p(m,n).$$ By assumption, the first formula in the chain is true, and each formula implies the next, so the last one is true.

It is possible to convert the rule into one analogous to "strong induction", which is more cumbersome to state. It weakens the inductive assumption to say that $p(u+1, v)$ and $p(u,v+1)$ must hold if $p(x,y)$ holds for all $x \leq u$ and $y \leq v$.