SOLUTION: Prove the following using indirect proof: 1. (A ∨ B) ⊃ C 2. (∼A ∨ D) ⊃ E / C ∨ E

Question 1193421: Prove the following using indirect proof:
1. (A ∨ B) ⊃ C
2. (∼A ∨ D) ⊃ E / C ∨ E

Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Here's how I'd do the derivation table

Number		Statement	Line(s) Used	Reason
1		(A v B) -> C
2		(~A v D) -> E
	:.	C v E
	3	~(C v E)		Assumption for Indirect Proof
	4	~C & ~E	3	De Morgan’s Law
	5	~C	4	Simplification
	6	~E	4	Simplification
	7	~(A v B)	1,5	Modus Tollens
	8	~A & ~B	7	De Morgan’s Law
	9	~A	8	Simplification
	10	~(~A v D)	2,6	Modus Tollens
	11	~~A & ~D	10	De Morgan’s Law
	12	A & ~D	11	Double Negation
	13	A	12	Simplification
	14	A & ~A	13,9	Conjunction
15		C v E	3-14	Indirect Proof, aka proof by contradiction

Further explanation:
In place of the horseshoe symbols, I used arrows.

The idea is to negate the conclusion C v E to get ~(C v E) as the starting point of the indirect proof or proof by contradiction (this is done on line 3).
The goal is to try to find two statements from that starting point that clash with one another.
That contradiction will then lead back to C v E being the only possibility.

On lines 9 and 13, I got ~A and A respectively. One or the other must happen, but both cannot be simultaneously the case.
It's like saying a light switch is off and on at the same time.
This is the key contradiction needed to conclude that ~(C v E) cannot be the case. Therefore, C v E is a valid conclusion.

Notice how the "number" column has been indented for the subcase of assuming the ~(C v E). It's like going off on a separate hypothetical branch. Once we realize a contradiction happens with that branch, we then snap back to the opposite of ~(C v E) in other words C v E

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