Question #349473

1.Show that each of these conditional statements is a tautology by using truth tables.

a) (p∧q)→ p

b) p → (p∨q)

c) ¬p → (p → q)

d) (p∧q)→ (p → q)

e) ¬(p → q)→ p

f) ¬(p → q)→¬q


2.Refer to 1, Show that each of these conditional statements is a tautology by using Propositional Logic


1
Expert's answer
2022-06-13T10:22:27-0400

1.

a) (p∧q)→ p




b) p → (p∨q)




c) ¬p → (p → q)




d) (p∧q)→ (p → q)




e) ¬(p → q)→ p




f) ¬(p → q)→¬q






2.

a) (pq)p=¬(pq)p=(¬p¬q)p=p¬p¬q=1¬q=1(p\wedge q)\to p=\neg(p\wedge q)\vee p=(\neg p\vee \neg q)\vee p=p\vee \neg p \vee \neg q=1\vee \neg q=1


b) p(pq)=¬p(pq)=¬ppq=1q=1p\to (p\vee q) = \neg p \vee(p\vee q)=\neg p\vee p\vee q=1\vee q=1


c) ¬p(pq)=¬¬p(pq)=p(¬pq)=p¬pq=1q=1\neg p\to(p\to q) =\neg\neg p \vee (p\to q)= p\vee (\neg p \vee q)=p\vee\neg p\vee q=1\vee q=1


d) (pq)(pq)=¬(pq)(pq)=(¬p¬q)(¬pq)=(p\wedge q)\to(p\to q)=\neg(p\wedge q)\vee(p\to q)=(\neg p\vee \neg q)\vee(\neg p\vee q)=

=¬p¬q¬pq=¬p¬qq=¬p1=1=\neg p\vee \neg q \vee \neg p\vee q = \neg p \vee \neg q\vee q = \neg p \vee 1=1


e) ¬(pq)p=¬¬(pq)p=(pq)p=¬pqp=1q=1\neg(p\to q)\to p =\neg\neg(p\to q)\vee p=(p\to q)\vee p=\neg p\vee q\vee p= 1 \vee q=1


f) ¬(pq)¬q=¬¬(pq)¬q=(pq)¬q=¬pq¬q=¬p1=1\neg(p\to q)\to\neg q =\neg\neg(p\to q)\vee\neg q=(p\to q)\vee \neg q=\neg p\vee q\vee\neg q=\neg p\vee 1=1



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