0

Suppose that I wanted to prove $P(n)$ is true, but I need some $Q(n)$ to be true so that I can show $P(n) \implies P(n+1)$. Is it always okay for me to do

$$P(n) \land Q(n) \implies P(n+1) \land Q(n+1)$$

to show that $P(n) \land Q(n)$ are true and then extract $P(n)$ ?

Tarang Saluja
  • 391
  • 6
    Yes, as long as you prove $P(0) \land Q(0)$. It occurs with some frequency in inductive proofs that it's easier to prove more than you want (because then you have more hypotheses to work with). – Robert Shore Sep 11 '20 at 19:49
  • 3
    Certainly. If you set $R(n) := P(n) \land Q(n)$ then your implication can be written $R(n) \implies R(n+1),$ which is on the ordinary form. – md2perpe Sep 11 '20 at 20:30

0 Answers0