Can anyone help me solve this (p → r) → (¬a v b), p → q, b → s, q → r, ¬a → s // (r v s)

Question

I have been working almost three days on this problem and I can't to this answer:

1. (p → r) → (¬a v b)
2. p → q
3. b → s
4. q → r
5. ¬a → s

// (r v s)

Answer 1

Hint

Assume p and derive r from 2. and 4. and then derive p → r by →-intro (Conditional Proof).

Now derive ¬a v b from 1. by →-elim (Modus Ponens) and use v-elim (Proof by Cases) to derive s.

Conclude with r v s by v-intro (addition).

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A different way (using Rules of inference) is:

A) apply Hypothetical syllogism to 2. and 4. to derive:

B) p → r

Apply Modus ponens to 1) and B) to derive:

c) ¬a v b

Apply Disjunction elimination to 3., 5. and C) to derive:

D) s

Finally, apply Addition to conclude with:

r v s.

Answer 2

Hopefully people will forgive my logic being slightly rusty, but here's my complete proof (I came up with it independently of @Mauro's answer but they end up being quite similar, except that mine has the details filled in):

1. (p → r) → (¬a v b)
2. p → q
3. b → s
4. q → r
5. ¬a → s

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6.a) Suppose p.
b) Then q (→ elimination, 6a and 2)
c) Then r (→ elimination, 6b and 4)
7. p → r (→ introduction, 6a-c)
8. (¬a v b) (→ elimination, 7 and 1)
9. Case analysis on (¬a v b)
a) Suppose ¬a
b) s (→ elimination, 9a and 5)
c) r v s (v introduction)
d) Suppose b
e) s (→ elimination, 9d and 3)
f) r v s (v introduction)
10. r v s (v elimination, 9a - 9f). QED