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Table10100Solutionoftheexample
p
0
0
0
0
1
1
1
1
q
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
r
p∨q
0
0
1
1
1
1
1
1
p⇔r
1
0
1
0
0
1
0
1
q∧r
0
0
0
1
0
0
0
1
(p∨q)∧(p⇔r)
0
0
1
0
0
1
0
1
Inthelaststep,wedeterminethelogicalvaluesofthewholeformula,bearingin
mindthatithasaformofimplication,withthecomponents(p∨q)∧(p⇔r)
andq∧r,andsothelogicalvaluesofthoseexpressionsmustbeusedinthe
calculation.Forexample,inthethirdrow,theyarerespectively1and0(notethe
order!),Sotheimplicationisfalse(Booleanvalue0),andinthefourthrow
0and1,sotheimplicationistrue(Booleanvalue1).Thefinalresultsare
presentedintheTable1.11.
Table10110Solutionoftheexample
0
0
0
0
1
1
1
1
p
0
0
1
1
0
0
1
1
q
0
1
0
1
0
1
0
1
r
p∨q
0
0
1
1
1
1
1
1
p⇔r
1
0
1
0
0
1
0
1
q∧r
0
0
0
1
0
0
0
1
(p∨q)∧(p⇔r)
0
0
1
0
0
1
0
1
(p∨q)∧(p⇔r)3(q∧r)
1
1
0
1
1
0
1
1
Inconclusion,afterexaminingtheeightpossiblecases,wefindthatintwoof
themtheformulaisfalse.Thismeansthatitisnotatautology(recall–tautology
isasentencealwaysthatistrue).
Example3
Itisknownthattheformulaep∨¬qandp3qaretrue.Whatcanwesayabout
thestatementspandq?
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