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Question 56 of 84

Q.Verify whether the statement q∨[p∨(∼q)]q \vee [p \vee (\sim q)] is a tautology or a contradiction.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 6mImportance★★★★★
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Construct the truth table over all four combinations of pp and qq and observe that the final column is always True.

  1. Statement: q∨[p∨(∼q)]q\vee[p\vee(\sim q)].

  2. Case p=T,q=Tp=T,q=T: ∼q=F\sim q=F; p∨(∼q)=T∨F=Tp\vee(\sim q)=T\vee F=T; q∨[…]=T∨T=Tq\vee[\ldots]=T\vee T=T.

  3. Case p=T,q=Fp=T,q=F: ∼q=T\sim q=T; p∨(∼q)=T∨T=Tp\vee(\sim q)=T\vee T=T; q∨[…]=F∨T=Tq\vee[\ldots]=F\vee T=T.

  4. Case p=F,q=Tp=F,q=T: ∼q=F\sim q=F; p∨(∼q)=F∨F=Fp\vee(\sim q)=F\vee F=F; q∨[…]=T∨F=Tq\vee[\ldots]=T\vee F=T.

  5. Case p=F,q=Fp=F,q=F: ∼q=T\sim q=T; p∨(∼q)=F∨T=Tp\vee(\sim q)=F\vee T=T; q∨[…]=F∨T=Tq\vee[\ldots]=F\vee T=T.

  6. Truth table summary:

    | pp | qq | ∼q\sim q | p∨(∼q)p\vee(\sim q) | q∨[p∨(∼q)]q\vee[p\vee(\sim q)] | …

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