Determine whether the statement (p∧(p⟶q))⟶q is a tautology one time by using truth table and the other time without using truth table

It is.........................................................

One way to prove this without a truth table is to use a conditional proof. We assume the portion p ^ (p --> q). If that's true, then so is p and p-->q

Using p and p-->q, the modus ponens rule allows us to derive q. It says that if p is true and p --> q, then q must be true as well.

Since we arrive at q, we have found the conclusion we're after. The assumption (p ^ (p-->q)) leads to q, and therefore the entire statement (p ^ (p-->q)) --> q is true for any combination of p,q.

step-by-step explanation:

once you graph it hit 2nd (the blue button) and graph

that takes you to a table of coordinates

on the left is x coordinates and on the right is y coordinates and an empty box

(you may have to scroll up/down in order to get ( x, 0)