Is it a correct solution ??

13 votes

Show that the conclusion $(r \to q)$ follows from the premises:

$p, (p \to q) \vee (p \wedge (r \to q))$

$p, (p \to q) \vee (p \wedge (r \to q))$

15 votes

Best answer

$P_1 :p$

$P_2: (p → q) ∨ (p ∧ (r → q) )$

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

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

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

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

$\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.

$P_2: (p → q) ∨ (p ∧ (r → q) )$

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

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

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

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

$\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.

1

[*P*∧[¬*P*∨*Q*∨(*P*∧¬*R*)∨(*P*∧*Q*)]]→(*R*→*Q*)

how did this become this [(*P*∧¬*R*)∨(*P*∧*Q*)]→(*R*→*Q*) please explain me this step

As this part *P*∧[(¬*P*∨*Q*) is modus ponens and should reduce to Q but Q is not in the wff

12 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)

=> (~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)

9 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).

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).

1 vote

If we look at it carefully it is given that if premises are true then if r is true then q must be true.

We have data p is true and r is true(as we want to prove it)

now we will try to prove that r implies q is not the solution. If you substitute p,r is true then truth value of second premise complete depend upon truth value of q. so it must be true and hence the conclusion.

We have data p is true and r is true(as we want to prove it)

now we will try to prove that r implies q is not the solution. If you substitute p,r is true then truth value of second premise complete depend upon truth value of q. so it must be true and hence the conclusion.

0 votes

Some n number of premises implies a conclusion means that, If all the n premises are true then conclusion is true as well.

p and (p→q)∨(p∧(r→q)) both premises will be true only for below truth value assignments

(p=T,r=F,q=T), (p=T,r=T,q=T), (p=T,r=F,q=F)

none of the above truth value assignments has r=T,q=F hence , If both the premises(p and (p→q)∨(p∧(r→q)) ) are true then (r→q) is true as well.( Note: (r→q) is false for only r=T,q=F truth value assignment)

p and (p→q)∨(p∧(r→q)) both premises will be true only for below truth value assignments

(p=T,r=F,q=T), (p=T,r=T,q=T), (p=T,r=F,q=F)

none of the above truth value assignments has r=T,q=F hence , If both the premises(p and (p→q)∨(p∧(r→q)) ) are true then (r→q) is true as well.( Note: (r→q) is false for only r=T,q=F truth value assignment)