Let T denotes True and F denotes False.
1.Let us determine whether ¬(p∨(¬p∧q)) and ¬p∧¬q are equivalent:
¬(p∨(¬p∧q))=¬p∧¬(¬p∧q))=¬p∧(p∨¬q))=
=¬p∧p∨¬p∧¬q=F∨¬p∧¬q=¬p∧¬q .
Therefore, the formulas are equivalent.
2.Let us determine whether the compound proposition ∼(p∨q)∨(∼p∧q)∨p is tautology.
∼(p∨q)∨(∼p∧q)∨p=∼(p∨q)∨(∼p∨p)∧(q∨p)=∼(p∨q)∨T∧(q∨p)=
=∼(p∨q)∨(q∨p)=∼(p∨q)∨(p∨q)=T.
Therefore, the formula ∼(p∨q)∨(∼p∧q)∨p is tautology.
3.Let us determine whether (p → q)∧(p → r) ≡ p → (q ∧r) using a truth table:
pFFFFTTTTqFFTTFFTTrFTFTFFFTp→qTTTTFFTTp→rTTTTFTFT(p→q)∧(p→r)TTTTFFFTq∧rFFFTFFFTp→(q∧r)TTTTFFFT
Since the formulas (p→q)∧(p→r) and p→(q∧r) always have the same truth values, they are logically equivalent.
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