Show that each of these conditional statements is a tautology by using truth tables. a) [¬p ∧ (p ∨ q)] → q b) [(p → q) ∧ (q → r)] → (p → r) c) [p ∧ (p → q)] → q d) [(p ∨ q) ∧ (p → r) ∧ (q → r)] → r

Step-by-step explanation:

To show that a statement is a tautology using truth table - is to show that all the entries in the expression are truths T.

We can do this by taking each statement, expression by expression. For example, to show that

[~p ∧ (p ∨ q)] → q

is a tautology, knowing we have 3 columns, we have 2^3 = 8 rows. We start by putting putting truth values for p, q, and r respectively

Next, we find ~p, then find (p ∨ q), then find ~p ∧ (p ∨ q), before finally arriving at the required [~p ∧ (p ∨ q)] → q

TERMINOLOGIES AND SYMBOLS

- T means Truth

- F means False

- ~p means negation of p.

~p is F if p is T, and vice versa

- ∧ means conjunction.

p ∧ q is T only if p is T and q is T.

- ∨ means disjunction.

p ∨ q is T if either p or q is T.

- → is for conditional 'if then'

p → q is T if both p and q are T, or both p and q are F, or p is F and q is T, otherwise, it is F.

THE STEP BY STEP WORKINGS FOR THE STATEMENTS GIVEN ARE IN THE ATTACHMENT.

4x and 6y

im not sure though

no solution

step-by-step explanation:

there is not enough information in the problem to solve.