Can modus tollens be statement of proof by contradiction or is it just a specific case of contradiction?
i.e we know that in general, proof by contradiction stated as follows
$[P' \implies (q \land q')] \implies P$
And by modus tollens, we have
$[(P' \implies q) \land q'] \implies P$
Here we assume $P'$ true and show q' happens, which should not happen: a contradiction. I tend to think modus tollens is foundation of proof by contradiction, but it seems just a specific case of contradictions...
On
The two are equivalent.
For example, having proof by contradiction we get :
1) $(\lnot P \to Q) \land \lnot Q$ --- premise
2) $\lnot P \to Q$ --- from 1) by conjunction-elimination
3) $\lnot Q$ --- from 1) by conjunction-elimination
4) $\lnot P$ --- assumed [a]
5) $Q$ --- from 2) and 4) by modus ponens
6) $Q \land \lnot Q$ --- from 3) and 5) by conjunction-introduction
7) $\lnot P \to (Q \land \lnot Q)$ --- from 4) and 6) by conditional proof, discharging [a]
8) $P$ --- from 7) and proof by contradiction and modus ponens
9) $(\lnot P \to Q) \land \lnot Q \vdash P$ --- from 1) and 8).
Actually, these are two different views of the same thing. In one case you begin with q' already proven, and prove P' -> q; then you apply the logical equivalence P' -> q <==> q' -> P, and use modus-ponens on the latter (plus q') to arrive at P. In the other case you prove P' -> (q^q'), apply the same equivalence to infer (q^q')' -> P, observe that (q^q')' is a tautology hence true, and apply modus-ponens to infer P.
And in principle you can convert the second form to the first: First prove P' -> q, introduce the tautology q' -> (P' -> q'), apply modus-ponens to that (plus 'q') to infer P' -> q', then combine with the first to derive P' -> (q^q').
It is really a question of what you have and what you need. In the first case you have neither q nor q' and so have to prove the contradiction from P' in its entirety. In the second case you have HALF of the contradiction from other sources, so you only need to prove the other half from P'.