How to prove that $\forall x( x + 0 = 0 + x)$ in Peano Arithmetic with Natural Deduction

Question

I was wondering if someone could give me a hint as to how to prove

$$\forall x( x + 0 = 0 + x)$$

in Peano Arithmetic using Natural Deduction. I can use

$$\forall x (x + 0 = x) \ \ \ \ \text{ and } \ \ \ \ \forall x( 0 + x = x),$$

a combination of which should intuitively get me the result. The issue is that the $\forall$-elimination rule

$$\frac{\begin{matrix}\vdots\\ \forall x\phi \end{matrix}}{\phi[t/x]} $$

allows me to substitute a variable by a term, and what I'd like is to substitute a variable $(x)$ by an open formula $(0 + x)$

Answer 1

how to prove $∀x(x+0=0+x)$?

In order to prove commutativity of sum, we have to use symmetry and transitivity of equality.

In $(\forall \text E)$ rule, the "instantiation" operation $\phi[t/x]$ means exactly that we have to replace the variable $x$ occurring into formula $\phi$ with a term $t$ whatever.

A term can be: a variable, a constant, like $0$, or a "complex" term built up using function symbols, like $0+0$ or $x+y$.

In your case, we have the formula $\forall x (x+0=x)$ and thus $(x+0=x)$ is the $\phi$.

We will apply $(\forall \text E)$ instantiating $\forall x (x+0=x)$ with $(x+0=x)$ itself or using another variable: $(y+0=y)$, $(a+0=a)$ etc.

Thus, from $\forall x (x+0=x)$ we get $(a+0=a)$ and from $\forall x (0+x=x)$ we get $(0+a=a)$.

Using symmetry of equality (see your previous post) from $(0+a=a)$ we get $(a=0+a)$.

The result follows from $(a+0=a)$ and $(a=0+a)$ using transitivity of equality and $(\forall \text I)$.