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Click and drag the steps in the correct order to show that (p V q) A (PV)-(qV) is a tautology. Cale But if q and rare both true, then one of p v q or "pvr is true, because one of p or p is true. Hence, the conditional statement is true if the hypothesis is false or the conclusion is true. The conclusion q v r will be true in every case except when q and rare both false. Thus, in this case, the hypothesis (pv q) ^ (pvr) is true. Thus, in this case, the hypothesis (p v q) A (p vr) is false. But if g and rare both false, then one of p v q or pvr is false, because one of p or p is false. The conclusion q v r will be true in every case except when q and r are both true. Reset

Click and drag the steps in the correct order to show that (p V q) A (PV)-(qV) is a tautology. Cale But if q and rare both true, then one of p v q or "pvr is true, because one of p or p is true. Hence, the conditional statement is true if the hypothesis is false or the conclusion is true. The conclusion q v r will be true in every case except when q and rare both false. Thus, in this case, the hypothesis (pv q) ^ (pvr) is true. Thus, in this case, the hypothesis (p v q) A (p vr) is false. But if g and rare both false, then one of p v q or pvr is false, because one of p or p is false. The conclusion q v r will be true in every case except when q and r are both true. Reset

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Click and drag the steps in the correct order to show that (p V q) A (p Vn) → (9V) is a tautology. But if q and rare both true, then one of pv qor pvr is true, because one of p or p is true. Hence, the conditional statement is true if the hypothesis is false or the conclusion is true. The conclusion q v r will be true in every case except when q and rare both false. Thus, in this case, the hypothesis (p v q) ^ (p v r) is true. Thus, in this case, the hypothesis (p v q) A (p v r) is false. But if q and rare both false, then one of p v q or p v r is false, because one of p or p is false. The conclusion q v r will be true in every case except when q and rare both true. Reset
Transcribed Image Text:Click and drag the steps in the correct order to show that (p V q) A (p Vn) → (9V) is a tautology. But if q and rare both true, then one of pv qor pvr is true, because one of p or p is true. Hence, the conditional statement is true if the hypothesis is false or the conclusion is true. The conclusion q v r will be true in every case except when q and rare both false. Thus, in this case, the hypothesis (p v q) ^ (p v r) is true. Thus, in this case, the hypothesis (p v q) A (p v r) is false. But if q and rare both false, then one of p v q or p v r is false, because one of p or p is false. The conclusion q v r will be true in every case except when q and rare both true. Reset
Prove the given expression is a tautology by developing a series of logical equivalence to demonstrate that it is logically equivalent to T. [pA (p→q)] →q DO r (pvq) v q by De Morgan's law pv (qv q) by associative law (p^q) →q=(pq) v q by logical equivalence (p^ q) →q by identity law (p^ q) →q=[-p v (p→q)] v q by logical equivalence [pA (p→q)] → q [p^ (p→q)] →q = [p^ (pv q)] →q by logical equivalence T by domination law pv (qv q) by associative law [F v (p ^ q)] →q by negation law [(p v p) ^ (pvq)] v q by distributive law < Prev ***** 11 of 32 Next > mai
Transcribed Image Text:Prove the given expression is a tautology by developing a series of logical equivalence to demonstrate that it is logically equivalent to T. [pA (p→q)] →q DO r (pvq) v q by De Morgan's law pv (qv q) by associative law (p^q) →q=(pq) v q by logical equivalence (p^ q) →q by identity law (p^ q) →q=[-p v (p→q)] v q by logical equivalence [pA (p→q)] → q [p^ (p→q)] →q = [p^ (pv q)] →q by logical equivalence T by domination law pv (qv q) by associative law [F v (p ^ q)] →q by negation law [(p v p) ^ (pvq)] v q by distributive law < Prev ***** 11 of 32 Next > mai
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