Question #187923

Construct a truth table for each of these compound propositions.

a) p ∧ ¬p

b) p ∨ ¬p

c) (p ∨ ¬q) → q

d) (p ∨ q) → (p ∧ q)

e) (p → q) ↔ (¬q → ¬p)

f) (p → q) → (q → p)


Expert's answer

Given:

a) p∧¬p\ p \wedge \lnot p

b) p∨¬p\ p \lor\lnot p

Answer:

p¬pp∧¬pp∨¬p01011001\begin{array}{ |c| c| c|c |} \hline p &\lnot p & p \wedge\lnot p& p \lor\lnot p \\ \hline 0 & 1 & 0&1\\ \hline 1 & 0 & 0&1 \\ \hline \end{array}

c) (p∨¬q)→q(p \lor\lnot q) \to q

d) (p∨q)→(p∧q)(p \lor q) \to (p \wedge q)

e) (p→q)↔(¬q→¬p)(p \to q) \leftrightarrow (\lnot q \to \lnot p)

f) (p→q)→(q→p)(p \to q) \to (q \to p)

Answer:

pq¬p¬qp∨¬q(p∨¬q)→qp∨qp∧q(p∨q)→(p∧q)p→q¬q→¬pq→p(p→q)↔(¬q→¬p)(p→q)→(q→p)00111000111111011001100110101001101000011111001111111111\begin{array}{ |c| c| c|c |c|c|c|c|c|c|c|c|c|c|} \hline p &q & \lnot p &\lnot q & p \lor\lnot q&(p \lor\lnot q) \to q & p \lor q& p \wedge q & (p \lor q) \to (p \wedge q) &p\to q&\lnot q \to \lnot p&q\to p&(p \to q) \leftrightarrow (\lnot q \to \lnot p)&(p \to q) \to (q \to p)\\ \hline 0 & 0 & 1&1&1&0&0&0&1&1&1&1&1&1\\ \hline 0 & 1 & 1&0&0&1&1&0&0&1&1&0&1&0 \\ \hline 1 & 0 & 0&1&1&0&1&0&0&0&0&1&1&1\\ \hline 1 & 1 & 0&0&1&1&1&1&1&1&1&1&1&1 \\ \hline \end{array}


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