Proof: $a = b \vdash f(a) = f(b)$

Question

How can I show: $a = b \vdash f(a) = f(b)$, preferable in natural deduction.

I am allowed to use the following rules:

• all for propositional logic ($\land_i, \land_e, \lor_i, \lor_e, \rightarrow_i, \rightarrow_e, \lnot_i, \lnot_e, \lnot \lnot i, \lnot \lnot e$
• predicate logic: $=_i, =_e, \forall_i, \forall_e, \exists_i, \exists_e$

Answer 1

Not sure what all your abbreviations mean, but here is how I would do it using standard logic. Assuming $f, a$ and $b$ had already been introduced...

1. Suppose $a=b$

2. $f(a)=f(a)$ by symmetry of equality

3. $f(a) = f(b)$ by substitution

4. We conclude: $a=b \implies f(a)=f(b)$