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Please help me create a truth table for the statement p^(qv~p)

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4 votes
Hello,

1) p∧(q∨¬p)=(p∧q)∨(p∧¬p)=p∧q proof:

2)

p|q|¬p|q∨¬p|p∧(q∨¬p)|p∧q
0|0|..1|.......1|..............0|0
0|1|..1|.......1|..............0|0
1|0|..0|.......0|..............0|0
1|1|..0|.......1|..............1|1



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