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Question Number 79974 by TawaTawa last updated on 29/Jan/20

Answered by Rio Michael last updated on 30/Jan/20

$$\mathbf{solution}$$$$[(p\vee q)\wedge (\sim p\vee r)]\Rightarrow (q\vee r)$$$$\mathrm{we}\mathrm{know}\mathrm{from}\mathrm{known}\mathrm{facts}\mathrm{that}$$$$-(p\vee q)\mathrm{is}\mathrm{a}\mathrm{contingency}\left(\mathrm{neither}\mathrm{a}\mathrm{tautology}\mathrm{nor}\mathrm{contradiction}\right)$$$$-(\sim p\vee r)\mathrm{is}\mathrm{a}\mathrm{contingency}$$$$\mathrm{so}(\mathrm{p}\vee q)\wedge (\sim p\vee r)\Rightarrow \mathrm{contingency}.$$$$\mathrm{When}\mathrm{ever}\mathrm{we}\mathrm{use}\mathrm{an}\mathrm{implication}(\Rightarrow )\mathrm{for}\mathrm{two}$$$${\mathrm{contingency}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{result}\mathrm{is}\mathrm{a}\mathrm{tautology}.$$$$\mathrm{But}(q\vee r)=contingency$$$$therefore[(p\vee q)\vee (\sim p\vee r)]\Rightarrow (q\vee r)isatautology.$$$$\mathbf{formally}.$$$$[(p\vee q)\wedge (\sim p\vee r)]\Rightarrow (q\vee r)=[(p\vee q)\wedge \sim (p\wedge r)]\Rightarrow (q\vee r)$$$$=\sim [(\sim p\wedge \sim q)\vee (p\wedge r)\Leftarrow (\sim q\vee \sim r)]$$$$=\sim \left(contradiction\right)$$$$=tautology$$$$$$

Commented by TawaTawa last updated on 03/Feb/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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