SOLUTION: Can you please help me solve this proof? I am stuck at line six. 1. (A → E) → (D ∨ C) 2. D → (~B → C) ∴ ~C

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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.

NumberStatementLines UsedReason
1(A -> E) -> (D v C)
2D -> (~B -> C)
:.~C -> (A v B)
3~CACP
4(D & ~B) -> C2EXP
5~(D & ~B)4,3MT
6~D v ~~B5DM
7~D v B6DN
8D -> B7MI
9~B -> ~D8Trans
10~(A -> E) v (D v C)1MI
11~(~A v E) v (D v C)10MI
12(~~A & ~E) v (D v C)11DM
13(A & ~E) v (D v C)12DN
14(D v C) v (A & ~E)13Comm
15[(D v C) v A] & [(D v C) v ~E]14Dist
16(D v C) v A15Simp
17(C v D) v A16Comm
18C v (D v A)17Assoc
19D v A18,3DS
20~~D v A19DN
21~D -> A20MI
22~B -> A9,21HS
23~~B v A22MI
24B v A23DN
25A v B24Comm
26~C -> (A v B)3-25CP


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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