I recently started learning Discrete Maths and currently studying rules of inference. I was looking at a proof of Hypothetical Syllogism, aka:
P→Q
Q→R∴ P→R
and I came across this proof of the above rule:
(1) P→Q (Hypothesis)
(2) Q→R (Hypothesis)
(3) P (Assumption)
(4) Q (1 and 3: Modus Ponens)
(5) R (2 and 4: Modus Ponens)
(6) P→R (3 – 5: if P, then R)
∴ ((P→Q) ∧ (Q→R)) → (P→R)
I understand all the steps, except for step (3). Why do we need to assume that P is true? In that case, isn't it just a conditional proof and no longer a tautology? If P is false, would't the proof no longer hold?
Best Answer
$(3), (4), (5)$ can be seen as a subproof, within the larger proof.
We assume $P$, in order to see what follows from $P$.
In this case, after two applications of modus ponens, we see that $R$ follows from that assumption, together with the hypotheses.
So we have proven if P, then R. (Which we state in line $(6)$. In symbols, we have proven $P \rightarrow R$. We haven't proven P. But we have proven P implies R. If P is false, the implication is true (any implication with a false antecedent is true). However, we know that if P is true, then so must be R.