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Solution to ((𝑝 β†’ π‘ž) ∧ (π‘ž β†’ π‘Ÿ)) β†’ (𝑝 β†’ π‘Ÿ) - Sikademy
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Archangel Macsika

((𝑝 β†’ π‘ž) ∧ (π‘ž β†’ π‘Ÿ)) β†’ (𝑝 β†’ π‘Ÿ)

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Let us construct the truth table of the following proposition.


f=((𝑝 β†’ π‘ž) ∧ (π‘ž β†’ π‘Ÿ)) β†’ (𝑝 β†’ π‘Ÿ)


\ \begin{array}{||c|c|c||c|c|c|c|c||} \hline\hline p & q & r &𝑝 β†’ π‘ž & π‘ž β†’ π‘Ÿ & (𝑝 β†’ π‘ž) ∧ (π‘ž β†’ π‘Ÿ) & 𝑝 β†’ π‘Ÿ & f\\ \hline\hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1\\ \hline 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1\\ \hline 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1\\ \hline 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1\\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline\hline \end{array}


It follows that the formulaΒ ((𝑝 β†’ π‘ž) ∧ (π‘ž β†’ π‘Ÿ)) β†’ (𝑝 β†’ π‘Ÿ)Β is a tautology.

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Question ID: mtid-5-stid-8-sqid-104-qpid-33