>
> Answers:
> Q1 : true
> Q2 : true
> Q3 : true
> Q4 : true
> Q5 : false
> Q6 : false
> Q7 : false
> Q8 : true
> ---------------
> And here's the detailed explanation. It assumes that a priori there is complete logical independence of the six fundamental statements used (i.e. without the three premises, there is no logical inconsistency in assuming any combination of truthhood or falsehood for each separately).
>
> Let the following letters represent statements:
>
> A = Bill takes the bus.
> B = The bus is late.
> C = Bill misses his appointment.
> D = Bill feels downcast.
> E = Bill should go home
> F = Bill gets the job.
>
> Then the 3 premises are:
>
> (1) (A and B) implies C.
> (2) (C and D) implies (not E).
> (3) (not F) implies (D and E).
> -----------------------
> Q1: (A and B) implies F?
>
> True:
> 1. The contrapositive of premise (2) is: E implies ((not C) or (not D))
> 2. Thus (D and E) implies (D and ((not C) or (not D))), which clearly implies (D and (not C)), which implies (not C)
> 3. Combining this with premise (3) gives: (not F) implies (not C).
> 4. Thus C implies F, and combined with premise (1) gives: (A and B) implies F.
> -----------------------
> Q2: (C and E) implies F?
>
> True:
> We already know C implies F, so therefore we know the weaker proposition (C and E) implies F.
> -----------------------
> Q3: This could be interpreted as either:
>
> Q3A: (B and (not F)) implies ((not A) or (not C))
>
> OR
>
> Q3B: (B implies (not A)) or ((not F) implies (not C))
>
> Q3A: True:
> 1. (B and (not F)) implies (B and D and E) by premise (3).
> 2. By step two of the reasoning for Q1, (D and E) implies (not C), so also (B and D and E) implies (not C).
> 3. Thus (B and (not F)) implies (B and D and E), which implies (not C), which implies ((not A) or (not C)).
>
> Q3B: True:
> 1. (C implies F) was shown in step 4 of Q1, so we have the contrapositive (not F) implies (not C), which is the second
> term in the "or" of Q3B. Thus it doesn't matter whether the first term is true (it is in fact false in some cases).
>
> I *think* that the interpretation they were asking was Q3B, by the way it was phrased.
> -----------------------
> Q4: (B and (not F)) implies (not A)?
>
> True:
> 1. The contrapositive of the true proposition Q1 is (not F) implies ((not A) or (not B)).
> 2. Thus (B and (not F)) implies (B and ((not A) or (not B))) which implies (B and (not A)) which implies (not A).
> -----------------------
> Q5: (not C) implies ((not E) and (not F))?
>
> False: consider the combination:
> A, B, C, F false;
> D, E true.
> These satisfy the premises but not Q5.
> -----------------------
> Q6: (B or C) implies D?
>
> False: consider the combination:
> A, C, D, E false;
> B, F true.
> These satisfy the premises but not Q6.
> -----------------------
> Q7: F implies ((not D) or (not E))
>
> False: consider the combination:
> A, B, C false;
> D, E, F true.
> These satisfy the premises but not Q7.
> -----------------------
> Q8: (E and A and B) implies (not D)?
>
> True:
> 1. (E and A and B) implies (E and C) by premise (1).
> 2. The contrapositive of premise (2) is (E implies ((not C) or (not D))).
> 3. Thus (E and A and B) implies (E and C) which implies (C and ((not C) or (not D))) which implies (C and (not D)) which
> implies (not D).
>
> --grem"Even if Q3B was the correct interpretation, Q3A is still the BETS GAEM"linn

After thinking about it, I realized I wasn't proving Q5, Q6, and Q7 to be false. I was actually proving "Q5 is provably true", "Q6 is provably true", and "Q7 is provably true" to be false.

There's a distinction there, because the propositions which are false are a subset of the propositions which are not provably true, and there's generally room in between (if not, the logical system is called "complete". In that case, every first-order logical proposition in the language is either provably true or provably false.)

Q5: (not C) implies ((not E) and (not F))?
True if A, B, C, E, F true; D false.
False if A, B, C, F false; D, E true.

Thus Q5 can not be proved true or false.

Q6: (B or C) implies D?
True if A, B, C, F false; D, E true.
False if A, C, D, E false; B, F true.

Thus Q6 can not be proved true or false.

Q7: F implies ((not D) or (not E))
True if A, B, C, F false; D, E true.
False if A, B, C false; D, E, F true.

Thus Q7 can not be proved true or false.

There are, of course, propositions which can be proved false. They are precisely the negations of the provably true propositions, for example, (not Q1), which would be:

not ((A and B) implies F), which reduces to (A and B and (not F)).

So the amended answers are:
Q1 : true
Q2 : true
Q3 : true
Q4 : true
Q5 : neither true nor false
Q6 : neither true nor false
Q7 : neither true nor false
Q8 : true

Here, I'm taking "true" to be synonymous with "provably true" and "false" to be synonymous with "provably false". It might actually be the convention for this type of problem to say "false" when you mean "not provably true", which is what I did the first time around.