Question

Choose the correct choice(s) regarding the following propositional logic assertion S:

$S:((P\wedge Q)\to R)\to ((P\wedge Q)\to (Q\to R))$

• A. S is a tautology .
• B. S is neither a tautology nor a contradiction.
• C. The antecedent of S is logically eqivalent to the consequent of S.
• D. S is a contradiction .

Answer 1

Answer: (A), (C)

The antecedent of S is logically eqivalent to the consequent of S.

$S:\left(\right(P\wedge Q)\to R)\to \left(\right(P\wedge Q)\to (Q\to R\left)\right)$
$\equiv (pq\to r)\to (pq\to (q\to r\left)\right)$
$\equiv \left[\right(pq){}^{\prime }+r]\to \left[\right(pq){}^{\prime }+(q{}^{\prime }+r)$]
$\equiv [pq\cdot r{}^{\prime }]+[p{}^{\prime }+q{}^{\prime }+q{}^{\prime }+r]$
$\equiv [pq\cdot r{}^{\prime }]+[p{}^{\prime }+q{}^{\prime }+q{}^{\prime }+r]$
$\equiv pqr{}^{\prime }+p{}^{\prime }+q{}^{\prime }+r$
$\equiv (p+p{}^{\prime })(qr{}^{\prime }+p{}^{\prime })+q{}^{\prime }+r$
$\equiv qr{}^{\prime }+p{}^{\prime }+q{}^{\prime }+r$
$\equiv (q+q{}^{\prime })(r{}^{\prime }+q{}^{\prime })+p^{\prime }+r$
$\equiv r{}^{\prime }+q{}^{\prime }+p{}^{\prime }+r\equiv r{}^{\prime }+r+q{}^{\prime }+p{}^{\prime }$
$\equiv 1+q{}^{\prime }+p{}^{\prime }\equiv 1$ (Tautology)
So, S is a tautology.
So, option (a) is true.
Option (b) and (d) are false.
Option (c) antecedent of S is
pq $\to r\equiv \left(pq\right){}^{\prime }+r$
$\equiv p{}^{\prime }+q{}^{\prime }+r$
The consequent of S is pq $\to (q\to r)$
$\equiv \left(pq\right){}^{\prime }+q{}^{\prime }+r$
$\equiv p{}^{\prime }+q{}^{\prime }+q{}^{\prime }+r$
$\equiv p{}^{\prime }+q{}^{\prime }+r$
SO, Antecedent of S $\equiv$ Consequent of S
So, option (c) is also true.