Propositional Logic with rules of inference problem. $$
\begin{array}{l}
1.\>\>\>\> (r ∧ ¬s) ∨ (q ∧ ¬s)\\
2.\>\>\>\> ¬s → ((p ∧ r) → u)\\
3.\>\>\>\> u → (s ∧ ¬t)\\
\dfrac{\quad\qquad\qquad\qquad\qquad\qquad}{\text{Prove from the previous arguments. }\quad p → q}
\end{array}
$$
Hey guys, I am really lost, so far I have a few arguments but not sure if they're correct. Can you please give some arguments to get this solved or a game plan using Inference laws and equivalences. Inferences: Modus Ponens, Tollens, Hypothetical, Disjunctive, Resolution, Conjunction, Simplification, Addition.
This is what I have so far....
4. (r ∨ q) ∧ ¬s Distributive of 1
5. ¬s Simplification of 1
6. (p ∧ r) → u Modus Ponens of 2
7. ¬u Modus Tollens of 3
8. ¬(p ∧ r) Modus tollens
9. p → ¬r DeMorgans and Implication Definition\
10. (r ∨ q) Simplification
11. ( ¬r → q) Implication
12. p → q Hypothetical
Are these steps valid?
 A: I'll give a summary "sketch" of one proof approach.
I'd suggest you start from the first premise: 
First check what follows from $r\land \lnot s$ and then what follows when $q \land \lnot s$. (At least one of these must be true. Why?)
From each of the above, we can use conjunction elimination (or simplification) to obtain $\lnot s$. And so, we can use disjunction elimination to conclude $\lnot s$.
From $\lnot s$ and premise $(2)$, we can derive (infer), by modus ponens, $(p\land r)\rightarrow u$. 
From $(p \land r) \rightarrow u$ together with premise $(3)$, we get $(p\land r) \rightarrow (s \land \lnot t)$
Now, we already derived $\lnot s.$ This means $\lnot s \lor t$ is true. Why?
But $\lnot s \lor t \equiv \lnot (s \land \lnot t).$
From $\lnot (s \land \lnot t),$ together with our derived $(p\land r )\rightarrow (s\land \lnot t)$, by modus ponens, we get $\not(p \land r)$
This means that either $\lnot p$, or $\lnot r$ is true.
If $\lnot p$ is true, we can build from this (addition) to get $\lnot p \lor q \equiv p\rightarrow q$, as desired.
If $\lnot r$ is true, then $r\land \lnot s$ is false (first alternative from premise $(1)$. Then it must follow that $q\land \lnot s$ is true. So it follows that $q$ is true (simplification). And given $q$, we have $\lnot p \lor q$ (addition), which is equivalent to $p \rightarrow q$, as desired.
Therefore, from $\lnot (p \land r)\equiv \lnot p \lor \lnot r$ we can conclude $(q\rightarrow q)$.
I'll leave this to you to write formally (including keeping proper track of temporary assumptions), with proper justifications.
A: Solution



*

*1:    (r∧¬s)∨(q∧¬s) Premise

*2:    ¬s→((p∧r)→u) Premise

*3:    u→(s∧¬t) Premise

*4:    (r∨q)∧ ¬s Distribution 1

*5:    ¬s Simplification 4

*6:    (p∧r)→u Modus Ponens 2, 5

*7:    ¬s∨t Addition 5

*8:    ¬(s∧¬t) DeMorgan 7

*9:    ¬u Modus Tollens 3, 8

*10:   ¬(p∧r) Modus Tollens 6, 9

*11:   ¬p∨¬r DeMorgan 10

*12:   p→¬r Material implication 11

*13:   r∨q Simplification 4

*14:   ¬r→q Material implication 13

*15:   p→q Hypothetical silogism 12, 14
