Answer to Question #349473 in Discrete Mathematics for mhist

Question

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

Answer 1

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) $(p\wedge q)\to p=\neg(p\wedge q)\vee p=(\neg p\vee \neg q)\vee p=p\vee \neg p \vee \neg q=1\vee \neg q=1$

b) $p\to (p\vee q) = \neg p \vee(p\vee q)=\neg p\vee p\vee q=1\vee q=1$

c) $\neg p\to(p\to q) =\neg\neg p \vee (p\to q)= p\vee (\neg p \vee q)=p\vee\neg p\vee q=1\vee q=1$

d) $(p\wedge q)\to(p\to q)=\neg(p\wedge q)\vee(p\to q)=(\neg p\vee \neg q)\vee(\neg p\vee q)=$

$=\neg p\vee \neg q \vee \neg p\vee q = \neg p \vee \neg q\vee q = \neg p \vee 1=1$

e) $\neg(p\to q)\to p =\neg\neg(p\to q)\vee p=(p\to q)\vee p=\neg p\vee q\vee p= 1 \vee q=1$

f) $\neg(p\to q)\to\neg q =\neg\neg(p\to q)\vee\neg q=(p\to q)\vee \neg q=\neg p\vee q\vee\neg q=\neg p\vee 1=1$