Question

Let p, q , r and s be four primitive statements . Consider the following arguments:
$P:\left[\right(\rightharpoondown p\vee q)\wedge (r\to s)\wedge (p\vee r\left)\right]\to (\rightharpoondown s\to q)$
$Q:\left[\right(\rightharpoondown p\wedge q)\wedge [q\to (p\to r)\left]\right]\to \rightharpoondown r$
$R:\left[\right[(q\wedge r)\to p]\wedge (\rightharpoondown q\vee p\left)\right]\to r$
$S:[p\wedge (p\to r)\wedge (q\vee \rightharpoondown r\left)\right]\to q$
Which of the above arguments are valid?

• A. P and Q only
• B. P and R only
• C. P and S only
• D. P,Q,R and S

Answer 1

The correct option is C P and S only

$P:\left[\right(\rightharpoondown p\vee q)\wedge (r\to s)\wedge (p\vee r\left)\right]\to (\rightharpoondown s\to q)$
$\equiv \left[\right(p\to q)\wedge (r\to s)\wedge (p\vee r\left)\right]\to (q\vee s)$
which is a rule of inference called constructive dilemma and therefore valid.

$S:[p\wedge (p\to r)\wedge (q\vee \rightharpoondown r\left)\right]\to q$
$\equiv p(p{}^{\prime }+r)(q+r{}^{\prime })\to q$
$\equiv pr(q+r{}^{\prime })\to q$
$\equiv prq\to q$
$\equiv \left(prq\right){}^{\prime }+q$
$\equiv p{}^{\prime }+r{}^{\prime }+q{}^{\prime }+q$
$\equiv p{}^{\prime }+r{}^{\prime }+1$
$\equiv 1$
Therefore S is valid .
Q and R can be similary simplified in boolean algebra to show that they are both not equivalent to 1 .
So only P and S are valid.
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