Let p, q, r and s be four primitive statements. Consider the following…

Check validity of each argument.

P: [((¬p∨q)∧(r→s)∧(p∨r))]→(¬s→q)

Valid. From:

• r→s

• ¬s

we get:

¬r

From:

p∨r and ¬r p

From: ¬p∨q and p, q

Hence: ¬s→q

So P is valid.

Q:[((¬p∧q)∧(q→(p→r)))]→¬r

From: ¬p, q and q→(p→r)

we get: p→r

But since p is false, p→r is automatically true regardless of r.

Cannot conclude ¬r.

So Q is invalid.

R: [((q∧r)→p)∧(¬q∨p)]→r

Take: q=F, p=F, r=F

Then:

• (q∧r)→p=T

• ¬q∨p=T

Premise true but conclusion r=F

Hence invalid.

S: [p∧(p→r)∧(q∨¬r)]→q

From:

• p

• p→r

we get: r

Now:

q∨¬r

Since r is true, ¬r is false.

Therefore, only P and S are valid.