# 11. Show that each conditional statement in Exercise 9 is a tautology without using truth tables

See below the truth table for a conditional statement, and conjunction and disjunctions of two propositions. We will use them to solve this exercise.

## a) (p∧q)→p

From the truth table for the conjunction, we know that if p^q is true, then p and q must be true.

If p is true, then (p^q)->p is also true as both sides are true.

If the hypothesis (p^q) is false, then (p^q)->p is true. Notice from the truth table for the conditional statement that if the hypothesis is false, then the conditional statement is true.

Therefore, (p^q)->p is a tautology.

## b) p→(p∨q)

If p is true, then pVq is also true (see the truth table for the disjunction -V- of two propositions).

If p is false, the conditional statement is true (truth table for the conditional statement).

Therefore, p→(p∨q) is a tautology.

## c) ¬p→(p→q)

If ¬p is true, then p is false. If p is false, p->q is always true (truth table for the conditional statement). It follows that ¬p→(p→q) is true (T->T is T).

If ¬p is false, then ¬p→(p→q) is true (truth table for the conditional statement).

Therefore, ¬p→(p→q) is a tautology.

## d) (p∧q)→(p→q)

If p^q is true, then p and q are also true (truth table for the conjunction of two propositions). It follows that p->q is true as both p and q are true (truth table for the conditional statement).

If p and q are true, then (p∧q)→(p→q) is also true (truth table for the conditional statement).

If p^q is false, then (p∧q)→(p→q)  is true (truth table for the conditional statement).

Therefore, (p∧q)→(p→q)  is a tautology.

## e) ¬(p→q)→p

If ¬(p→q) is true, then p->q is false.

According to the truth table for conditional statements, p->q is false only when p is true and q is false. Because p is true, it follows that ¬(p→q)→p is true.

If ¬(p→q) is false, then ¬(p→q)→p is true (truth table for the conditional statement).

Therefore, ¬(p→q)→p is a tautology.

## f) ¬(p→q)→¬q

If ¬(p→q) is true, then p->q is false.

According to the truth table for conditional statements, p->q is false only when p is true and q is false.

If q is false, then ¬q is true. It follows that ¬(p→q)→¬q is true.

If ¬(p→q) is false, then ¬(p→q)→ ¬q is true (truth table for the conditional statement).

Therefore, ¬(p→q)→¬q is a tautology.

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