tautology...?

Question

prove that [ (p v q) ^ (p -> r) ^ (q -> r) ] -> r is a tautology.

( the -> symbol is meant to be and arrow)

Answer 1

P=q ~^ r (56.1) and vice versa

Answer 2

To prove that

[ (p v q) ^ (p -> r) ^ (q -> r) ] -> r

is a tautology, all you have to do is create a truth table with p, q, and r, and prove that is true no matter what.

First, let's make our truth table.

p q r
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

Note that if r is true, the whole conditional is true, so we can deal with all of the other cases.

p q r
T T T = T {r is true here}
T T F
T F T = T {r is true here}
T F F
F T T = T {r is true here}
F T F
F F T = T {r is true here}
F F F

We have a conjunction as the antecedent, and if the antecedent is false, it follows that the conditional is true.
For a conjunction to be false, *any* of the conjuncts must be false. Let's take the conjuncts one at a time.

(p v q): this is a disjunction, and both p and q have to be false for a disjunction to be false. So for the truth table where P = F and Q = F, we have true.

p q r
T T T = T {r is true here}
T T F
T F T = T {r is true here}
T F F
F T T = T {r is true here}
F T F
F F T = T {r is true here}
F F F = T {both p and q are false here}

For p -> r: This is true if r is false or p is true. Look up the table where r is false OR p is true, and we get a true statement.

p q r
T T T = T {r is true here}
T T F = T {r is false here}
T F T = T {r is true here}
T F F = T {r is false here}
F T T = T {r is true here}
F T F = T {r is false here}
F F T = T {r is true here}
F F F = T {both p and q are false here}

And, as you can see, it's true all the time. Therefore it's a tautology.

Answer 3

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