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23 votes
23 votes
Show that the conclusion $(r \to q)$ follows from the premises$:p, (p \to q) \vee (p \wedge (r \to q))$
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9 Answers

Best answer
19 votes
19 votes
$P_1 :p$

$P_2: (p → q) ∨ (p ∧ (r → q) )$

$P_1 \wedge P_2 \rightarrow   (r \rightarrow q)$

$\equiv [p \wedge [(\neg p \vee q) \vee (p\wedge \neg r \vee (p \wedge q))]  \rightarrow (r \rightarrow q)$
$\equiv [p \wedge [\neg p \vee q \vee  (p\wedge \neg r) \vee  (p\wedge q)]] \rightarrow (r \rightarrow q)$
$\equiv [ (p\wedge \neg r) \vee (p\wedge q)] \rightarrow (r \rightarrow q)$
$\equiv [ (\neg p\vee r )\wedge (\neg p\vee  \neg q)\vee \neg r \vee q] $
$\equiv [ \neg p\vee  (r\wedge \neg q) \vee \neg r \vee q]$
$\equiv [\neg p \vee \neg q \vee \neg r \vee q] $
$\equiv [\neg p \vee  (q \vee \neg q) \vee \neg r] $
$\equiv 1$ (Tautology)

Hence, proved.
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14 votes
14 votes
(p→q)∨(p∧(r→q))

=> (~p V q ) V (p ∧ (~r V q))

=> (~p V q) V ( (p∧ ~r) V (p ∧ q) )

=> ( 0 V q ) V ( (1∧ ~r) V (1 ∧ q))

=> q V ( ~r V q )

=> ( ~r V q )

=> (r->q)
11 votes
11 votes
Using Distributive law, (p→q) ∨ (p ∧ (r→q)) = ((p→q) ∨ p) ∧ ((p→q) ∨ (r→q))

Using Simplification, (p→q) ∨ (r→q) is a conclusion.

(p→q) ∨ (r→q) = (¬p ∨ q) ∨ (¬r ∨ q) = ¬p ∨ q ∨ ¬r = ¬p ∨ (r→q)

Using premises p and ¬p ∨ (r→q) and applying Disjunction Syllogism, the conclusion is (r→q).
4 votes
4 votes
solution
To derive the conclusion we must assume that the premises are true

 

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