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Hat (π β§ π)β‘T, must be that πβ‘T and πβ‘T, hence [(π β§ π) β π]β‘ π, and [[(π β§ π) β π] β π ]β‘ π β Fβ‘Β¬π. Secondly, [(π β π) β π ] from t
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Therefore proving that [[(π β§ π) β π] β π ] β [(π β π) β π ] β‘ T. Q.E.DTherefore proving that [[(π β§ π) β π] β π ] β [(π β π) β π ] β‘ T. Q.E.DTherefore proving that [[(π β§ π) β π] β π ] β [(π β π) β π ] β‘ T. Q.E.D
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herefore π β¨ (π β¨ π)β‘F. So for the left side, since no matter if it is πβ‘F or
(π β¨ π)β‘F, one out of the two possibility being false is certain, so necessarily π β§ (π β¨ π)β‘F, and since F Fβ‘T, [π β§ (π β¨ π)] [(π β§ π) β¨ (π β§ π)] is true.
And consider the second case where (π β§ π) β¨ (π β§ π)β‘T, for it to be held true, necessarily (π β§ π)β‘T or (π β§ π)β‘T. Hence either π β§ (π β§ π) β‘T or π β§ (π β§ π) β‘F. Thus for π β§ (π β¨ π), its truth value must be T.
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