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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

3 votes
3 votes
The conclusion (r→q) follows from the premises :p, (p→q)∨(p∧(r→q)) means if the premises are true then conclusion must be true or we can say that it must be a valid argument .

Let us make conclusion false and then try to make premises true . If we are able to make conclusion false and premises true then it is not a valid argument .

So , r->q = false means  r = True and q = False .

Now let us make premises true :

p = True

(p→q)∨(p∧(r→q)) = True , that means either p->q = T or (p∧(r→q)) = T .

Substituting p =T , q= F and r = T in this premise we get :

p->q = T -> F = False

(p∧(r→q)) = (T∧(T→F)) = T∧F = False

In Neither case , we are able to make this premise (p→q)∨(p∧(r→q)) as True .

Since , we are not able to make conclusion false and premises true therefore it is a valid argument i.e if the premises are true then conclusion must be true .
Hence , The conclusion (r→q) follows from the premises :p, (p→q)∨(p∧(r→q))
1 votes
1 votes
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.
0 votes
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)
0 votes
0 votes
Can be done intuitively as follows:
We are given $p$ and we are given $(p→q)∨(p∧(r→q))$

If $p$ is true and $(p→q)∨(p∧(r→q))$ is true, it implies that either $q$ is true or $(r→q)$ is true.

But $(r→q)$ is true means $\lnot r \lor q$ is true

So,

$$p \land (p→q)∨(p∧(r→q))$$
$$\implies q \lor (r→q)$$
$$ \equiv q \lor \lnot r \lor q $$
$$\equiv \lnot r \lor q $$
$$\equiv (r\rightarrow q)$$

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