Using natural deduction to prove P ∧ ¬Q ∧ (R → S), ¬(P → S) ⊢ R → Q

Question

I would like to prove:

$1. \qquad P ∧ ¬Q ∧ R → S \qquad (Premise)$

$2. \qquad ¬(P → S) \qquad (Premise)$

$ ...$

$3. \qquad R → Q \qquad$

with access to these rules (http://imgur.com/kPZEYtG) However I am not sure how to proceed after this step:

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As a side note does anyone know how to enter $P ∧ ¬Q ∧ R → S, ¬(P → S) ⊢ R → Q$ into this great online natural deduction with steps tool I am not sure how to enter premises.

Answer 1

Here is a straightforward proof:

Answer 2

Lets prove it by contradiction.

We will try to make the Premises as True and Conclusion as False. (If we are not able to do this, then the inference is True)

To make the conclusion R→Q false, we must take R as True and Q as False (no other way we can make it false).

Considering these values of R and Q let us try to make the premises as True.

Premise 1: P ^ -Q ^ R→S

We have to take R as True and Q as false. So, P ^ -Q (True) ^ R (True)→S

This expression can be made True only by putting P as True and S as True. If any of these are False, Premise 1 becomes False.

Premise 2: -(P→S)

We have already taken P as True and S as True. So Premise 2 becomes False.

Therefore there is no such case when the Premises are True and Conclusion is False.

Hence the Inference is True. Proved.

Hope it helped.