Question 1009274: Can you please help me solve this proof? I am stuck at line six.
1. (A → E) → (D ∨ C)
2. D → (~B → C) ∴ ~C → (A ∨ B)
|3. ~C Assume
||4. ~A Assume
||5. (D∙~B)→C 2, EX
||6.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The idea is to assume ~C is true (line 3). Using the rules of inference/replacement, if we can lead to A v B somehow (line 25), then that proves ~C -> (A v B) is true.
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | (A -> E) -> (D v C) | | | | 2 | | D -> (~B -> C) | | | | :. | | ~C -> (A v B) | | | | 3 | ~C | | ACP | | 4 | (D & ~B) -> C | 2 | EXP | | 5 | ~(D & ~B) | 4,3 | MT | | 6 | ~D v ~~B | 5 | DM | | 7 | ~D v B | 6 | DN | | 8 | D -> B | 7 | MI | | 9 | ~B -> ~D | 8 | Trans | | 10 | ~(A -> E) v (D v C) | 1 | MI | | 11 | ~(~A v E) v (D v C) | 10 | MI | | 12 | (~~A & ~E) v (D v C) | 11 | DM | | 13 | (A & ~E) v (D v C) | 12 | DN | | 14 | (D v C) v (A & ~E) | 13 | Comm | | 15 | [(D v C) v A] & [(D v C) v ~E] | 14 | Dist | | 16 | (D v C) v A | 15 | Simp | | 17 | (C v D) v A | 16 | Comm | | 18 | C v (D v A) | 17 | Assoc | | 19 | D v A | 18,3 | DS | | 20 | ~~D v A | 19 | DN | | 21 | ~D -> A | 20 | MI | | 22 | ~B -> A | 9,21 | HS | | 23 | ~~B v A | 22 | MI | | 24 | B v A | 23 | DN | | 25 | A v B | 24 | Comm | | 26 | | ~C -> (A v B) | 3-25 | CP |
Acroynyms/Abbreviations used
ACP = assumption for conditional proof
Assoc = associative property
Comm = commutation
CP = conditional proof
Dist = distribution
DM = de morgan's law
DN = double negation
DS = disjunctive syllogism
EXP = exportation
HS = hypothetical syllogism
MI = material implication
MT = modus tollens
Simp = simplification
Trans = transposition
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