( ( [(p/\q)/\r] \/ [(p/\q)/\~r]) \/~q ) -> s how can we simplify the compound statement? please help me I must do it until monday I can do up to three steps but ı dont know what it can continue (Headbang)(Headbang)(Headbang)

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- Nov 1st 2008, 12:20 PMlozgurhow can we simplify ? pfff:(
( ( [(p/\q)/\r] \/ [(p/\q)/\~r]) \/~q ) -> s how can we simplify the compound statement? please help me I must do it until monday I can do up to three steps but ı dont know what it can continue (Headbang)(Headbang)(Headbang)

- Nov 1st 2008, 04:56 PMvagabond
You might need to use the distributive law for disjunction and conjunction .

For example, (A/\B)\/(A/\~B) = A/\(B\/~B) = A

Additionally, A->B equals ~A\/B (ex.((A->B) <-> (~A\/B))).

After simplifying your formula, you might need to verify the result using a truth table. - Nov 1st 2008, 05:34 PMSoroban
Hello, lozgur!

Vagabond is absolutely correct.

I'll run though the steps and hope you know the reasons.

Quote:

Simplify: .$\displaystyle \bigg\{\bigg[(p \wedge q \wedge r) \vee (p \wedge q \wedge \sim r)\bigg] \vee\sim q \bigg\} \to s $

Inside the brackets: .$\displaystyle (p \wedge q \wedge r) \vee (p \wedge q \:\wedge \sim r)$

"Factor": .$\displaystyle (p \wedge q) \wedge \underbrace{(r \:\vee \sim r)}$

. . . . . $\displaystyle = \;\underbrace{(p \wedge q) \quad\wedge\quad t}$

. . . . . $\displaystyle = \qquad\;\; p \wedge q$

The problem becomes: .$\displaystyle \bigg[(p \wedge q) \:\vee \sim q\bigg] \to s$

Distribute: . . . . $\displaystyle \bigg[(p \:\vee \sim q) \wedge (q \:\vee \sim q)\bigg] \to s$

. . . . . . . . . . . . . . . $\displaystyle \bigg[(p \:\vee \sim q) \wedge t\bigg] \to s $

. . . . . . . . . . . . . . . . . . $\displaystyle (p \:\vee \sim q) \to s$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

How far should we go? . . . There's no limit . . .

$\displaystyle (p \;\vee \sim q) \to s$

. . $\displaystyle \sim(p \;\vee \sim q) \vee s$

. . . $\displaystyle (\sim p \wedge q) \vee s$ . . . another answer

$\displaystyle (p \:\vee \sim q) \to s$

. . $\displaystyle (\sim q \vee p) \to s$

. . $\displaystyle (q \to p) \to s$ . . . and another

- Nov 2nd 2008, 01:55 AMlozgur
thank u very much friends for helping and sharing the solution (ı dont know advenced level english but ı know upper-intermeiated so maybe ı did some mistake writing ıam hoping u can understand. )

- Nov 2nd 2008, 02:24 AMlozgur
Question:

For each of these sets of premises, what relevant conclusion or conclu-

sions can be drawn? Explain the rules of inference used to obtain each

conclusion from premises.(a) "If I eat spicy foods, then I have strange dreams." "I have strange

dreams if there is thunder while I sleep." "I did not have strange

dreams."

(b) "I am dreaming or hallucinating." "I am not dreaming." "If I am

hallucinating, I see elephants running down the roads."

ı instaled this statement:

(a)

p:I eat spicy foods

q:I have strange dreams

r:there is thunder while I sleep

(p->q)/\(r->q)/\~q

do ı solve [(p->q)/\(r->q)/\~q] this statement for this question? ı dont understand what this question say

- Nov 2nd 2008, 03:46 AMlozgur
ı could do it (Clapping)(Clapping)