A4Q1(a)

[Jonathan Buss]

I'm confused by this question...
Isn't (a) an axiom? How should we go along to prove this?

It's an axiom, hence it is logically implied by the empty set under all
interpretations and valuations, and a valid formula.

Well, not quite. The argument, "It's an axiom, therefore it's valid",
relies on assumptions about the person who made it an axiom -- that s/he
meant it to be universally applicable, and s/he didn't make a mistake.

The question asks you to demonstrate that the statement is valid. That
demonstration forms part of the proof of the Soundness Theorem.

Think about how you might prove something like |= A -> (B -> A) without
soundness (since it's an axiom, the formal proof is a trivial one line).
The tutor who runs my tutorial section seems to like proofs by
contradiction, and that would probably be the easiest way to do it if
you're stuck.

"Proof by contradiction" indeed can work fine. Other approaches also
work fine.

Jonathan Buss

[Stephen Pisani]

I'm confused by this question...
Isn't (a) an axiom? How should we go along to prove this?

It's an axiom, hence it is logically implied by the empty set under all
interpretations and valuations, and a valid formula.

Think about how you might prove something like |= A -> (B -> A) without
soundness (since it's an axiom, the formal proof is a trivial one line).
The tutor who runs my tutorial section seems to like proofs by
contradiction, and that would probably be the easiest way to do it if
you're stuck.