False logic argument I have to proof the following theorem $$(w ⇒ p ∨ s) ∧ (t ∧ p ⇒ r) ⇒ (t ∧ ¬ r ⇒ ¬ q)$$
Using the assumptions, I would like to proof that $(t∧p⇒r)⇒(t∧¬r⇒¬q)$, but I got only $(p⇒r)⇒(t∧¬r⇒¬q)$. I thought I could use $(t∧p⇒r) ⇒ (p⇒r)$, but it seems to be false. How can I modify/finish this proof?
The answer of @DonAntonio is great, but I can't work with this. I would need a complete, linear proof.
 A: Shorter and probably  simpler proof by reductio ad absurdum: suppose that isn't true that $\;(t\wedge\neg r)\rightarrow\neg q\;$ . Then it must be that $\;t\wedge\neg r=T\;$  and $\;\neg q=F\iff q= T \;$ , and both $\;t=T\,,\,\neg r= T\iff r=F\;$
But then, since $\;(t\wedge p)\rightarrow r=T\;$ (assumption given), it must be that $\;t\wedge p=F\;$ . Since $\;t=T\;$ , this means that it must be $\;p=F\;$.
Also, we have the assumption $\;s\rightarrow\neg t=T\;$ . But $\;\neg t=F\;$ , so it must be that also $\;s=F\;$.
And we then get our contradiction with the first assumption, since $\;q=T\;$ yet $\;s\vee p=F\;$ as both $\;s= p= F\;$ .
A: I will gently disagree with DonAntonio: a linear proof does not have to be long and tiring, but it may very well be the way you do it (no offense, Leo!).
Your proof
First, let me address your proof. You need to make coarser steps. I don't know what class you're in (it looks like you are using perhaps a Gries and Schneider text), but please don't use contrapositive to turn $p \Rightarrow \neg q$ into $\neg\neg q \Rightarrow \neg p$ and then a separate step to replace $\neg\neg q$ with $q$. It is fine to have one form of contrapositive be simply: $p \Rightarrow \neg q \ \equiv \ q \Rightarrow \neg p$.
A similar comment applies to the steps of your calculation where you replace $s \Rightarrow \neg t$ with $\bf true$, then commute $\bf true$ to the other side of the formula for some reason, then eliminate it using the identity property of $\wedge$. This level of detail is not appropriate for the complexity of proof we are dealing with here.
If we clean up that mess, here is your calculation:

$\quad t \wedge \neg r \;\Rightarrow\; \neg q$
$\Leftarrow \quad \{$ using $q \Rightarrow s \vee p$ and monotonicity $\}$
$\quad t \wedge \neg r \ \Rightarrow \ \neg(s \vee p)$
$\equiv \quad \{$ de Morgan $\}$
$\quad t \wedge \neg r \ \Rightarrow \ \neg s \wedge \neg p$
$\Leftarrow \quad \{$ monotonicity $\}$
$\quad (t \Rightarrow \neg s) \;\wedge\; (\neg r \Rightarrow \neg p)$
$\equiv \quad \{$ contrapositive, twice $\}$
$\quad (s \Rightarrow \neg t) \;\wedge\; (p \Rightarrow r)$
$\equiv \quad \{$ using $s \Rightarrow \neg t$ $\}$
$\quad p \Rightarrow r$

The problem is, we only have $t \wedge p \Rightarrow r$, which is weaker than $p \Rightarrow r$, so this proof is simply not viable. The reason it is not is because of the third step, where you strengthened the calculandum into something too strong to prove.
My general suggestion to you is to be more careful with strengthening steps, and only use equivalence where possible.
A different approach
If you are calculating with booleans or anything else, there is a certain trade-off you have to be aware of. Either you do some work to "clean up" the given formulae, or you will have to do that work while calculating.
To be more specific, let me label the formulae in question:

$(0) \quad q \,\Rightarrow\, s \vee p$
$(1) \quad s \Rightarrow \neg t$
$(2) \quad t \wedge p \,\Rightarrow\, r$
$(3) \quad t \wedge \neg r \,\Rightarrow\, \neg q \quad.$

Now, if I don't wish to do any work up front, I might try to prove $(3)$ by strengthening $\neg q$ into $t \wedge \neg r$:

$\quad \neg q$
$\Leftarrow \quad \{$ using $(0)$ and antimonotonicity $\}$
$\quad \neg (s \vee p)$
$\equiv \quad \{$ de Morgan $\}$
$\quad \neg s \wedge \neg p$
$\Leftarrow \quad \{$ using $(1)$ and antimonotonicity $\}$
$\quad t \wedge \neg p$
$\Leftarrow \quad \{$ using $(2)$ in the form $t \wedge \neg r \,\Rightarrow\, \neg p$ and monotonicity $\}$
$\quad t \wedge t \wedge \neg r$
$\equiv \quad \{$ idempotence $\}$
$\quad t \wedge \neg r$

Personally, I think this is a bad proof: too much work is crammed into each step.
Cleaning up the givens
My taste is to instead do some work up front to clean up the given formulae. First, I use the contrapositive to rewrite $(3)$ as the equivalent:
$(3') \quad t \wedge q \,\Rightarrow\, r \quad.$
Given the shape of $(2)$:
$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r \quad,$
this suggests that we try to prove $(3')$ by strengthening $r$ into $t \wedge q$. We start as follows:

$\quad r$
$\Leftarrow \quad \{ \ (2) \ \}$
$\quad t \wedge p \quad.$

Now we need to strengthen $p$, so we rearrange $(0)$ as the equivalent:
$(0') \quad q \wedge \neg s \,\Rightarrow\, p \quad.$
Thus we can continue our calculation:

$\quad r$
$\Leftarrow \quad \{ \ (2) \ \}$
$\quad t \wedge p$
$\Leftarrow \quad \{ \ (0') \ \}$
$\quad t \wedge q \wedge \neg s \quad.$

Finally, we need to strengthen $\neg s$ so we rewrite $(1)$ as the equivalent:
$(1') \quad t \Rightarrow \neg s \quad,$
and finish the calculation as follows:

$\quad r$
$\Leftarrow \quad \{ \ (2) \ \}$
$\quad t \wedge p$
$\Leftarrow \quad \{ \ (0') \ \}$
$\quad t \wedge q \wedge \neg s$
$\Leftarrow \quad \{ \ (1') \ \}$
$\quad t \wedge q \wedge t$
$\equiv \quad \{$ idempotence $\}$
$\quad t \wedge q \quad.$

The cleanliness and clarity of each step suggests that this was a nice rewriting of the givens. We simply used shunting and contraposition to rewrite everything compatibly. For reference, we used the equations in the following forms:

$(0') \quad q \wedge \neg s \,\Rightarrow\, p$
$(1') \quad t \Rightarrow \neg s$
$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r$
$(3') \quad t \wedge q \,\Rightarrow\, r \quad.$

Note that because of the shape of $(1)$, it was inevitable that one of our symbols would need to be negated. In our case, we used $\neg s$. However, note that we never used $s$ without the negation symbol. Thus, letting $S = \neg s$ we could write everything as:

$(0') \quad q \wedge S \,\Rightarrow\, p$
$(1') \quad t \Rightarrow S$
$(2) \hphantom{'} \quad t \wedge p \,\Rightarrow\, r$
$(3') \quad t \wedge q \,\Rightarrow\, r \quad,$

and see that, by doing the work up front, negation no longer has anything to do with the structure of our argument.
The moral
The moral of the story here is that in any field of mathematical inquiry, it is important to massage our symbols and concepts so they are compatible with each other. Perhaps this is a simple problem from logic, but with a mess of symbols in front of you, even that simple problem confused you. How can we hope to prove complicated theorems in algebra, analysis, etc, if we can't even neatly tackle simple problems?
Be clean, be organized. It's no guarantee of a proof but it's the best we can do.
Appendix
If you are working in a rigid framework that doesn't allow you to do this sort of "cleaning" work up front, here is an alternate, "wide" calculation using the original formulae:

$(0) \quad q \,\Rightarrow\, s \vee p$
$(1) \quad s \Rightarrow \neg t$
$(2) \quad t \wedge p \,\Rightarrow\, r$
$(3) \quad t \wedge \neg r \,\Rightarrow\, \neg q \quad.$

Here is the calculation of $(3)$:

$\quad t \wedge \neg r \,\Rightarrow\, \neg q$
$\Leftarrow \quad \{$ using $(2)$ and $(0)$ and monotonicity $\}$
$\quad t \wedge \neg(t \wedge p) \,\Rightarrow\, \neg(s \vee p)$
$\equiv \quad \{$ de Morgan, twice $\}$
$\quad t \wedge (\neg t \vee \neg p) \,\Rightarrow\, \neg s \wedge \neg p$
$\equiv \quad \{$ absorption on the left side $\}$
$\quad t \wedge \neg p \,\Rightarrow\, \neg s \wedge \neg p$
$\Leftarrow \quad \{$ monotonicity of $\wedge$ $\}$
$\quad t \Rightarrow \neg s$
$\equiv \quad \{$ contrapositive $\}$
$\quad s \Rightarrow \neg t$
$\equiv \quad \{ \ (1) \ \}$
$\quad {\bf true} \quad.$

To clarify, I don't endorse this sort of calculation, generally speaking!
A: I'm putting my second reply to @DonAntonio into its own answer. If we take a substitution-of-values approach, the ideal candidate to me seems to be to assume $t$ — that is, to let $t = \bf true$.
Then our givens reduce to:

*

*$q \,\Rightarrow\, s \vee p$

*$\neg s \quad$ (that is, $s = \bf false$)

*$p \Rightarrow r$
and our demonstrandum reduces to:

*

*$\neg r \Rightarrow \neg q \quad.$
But from $s = \bf false$, the first given reduces to $q \Rightarrow p$, which in combination with $p \Rightarrow r$ yields $q \Rightarrow r$, which is just the contrapositive of what we're trying to show.
Of course, if $\neg t$ then our demonstrandum is vacuously true. (This was the reason we chose to perform a case analysis on $t$.) So the theorem is proved in any case.
