**Construct a truth tabal for each of these compound proposition**

The **Answer to the Question**

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**Here's the Solution to this Question**

Incomplete question:

We can take this example:

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)

Solution:

Given:

a)$\ p \wedge \lnot p$

b)$\ p \lor\lnot p$

Answer:

$\begin{array}{ |c| c| c|c |} \hline p &\lnot p & p \wedge\lnot p& p \lor\lnot p \\ \hline 0 & 1 & 0&1\\ \hline 1 & 0 & 0&1 \\ \hline \end{array}$

c) $(p \lor\lnot q) \to q$

d) $(p \lor q) \to (p \wedge q)$

e) $(p \to q) \leftrightarrow (\lnot q \to \lnot p)$

f) $(p \to q) \to (q \to p)$

Answer:

$\begin{array}{ |c| c| c|c |c|c|c|c|c|c|c|c|c|c|} \hline p &q & \lnot p &\lnot q & p \lor\lnot q&(p \lor\lnot q) \to q & p \lor q& p \wedge q & (p \lor q) \to (p \wedge q) &p\to q&\lnot q \to \lnot p&q\to p&(p \to q) \leftrightarrow (\lnot q \to \lnot p)&(p \to q) \to (q \to p)\\ \hline 0 & 0 & 1&1&1&0&0&0&1&1&1&1&1&1\\ \hline 0 & 1 & 1&0&0&1&1&0&0&1&1&0&1&0 \\ \hline 1 & 0 & 0&1&1&0&1&0&0&0&0&1&1&1\\ \hline 1 & 1 & 0&0&1&1&1&1&1&1&1&1&1&1 \\ \hline \end{array}$