I'm working on an assignment and I'm stuck on this proof. I feel like I'm on the right track but I can't find the way to prove the goal.
B ^ D
(B^¬A) → ¬C
B → ¬A
(D^E)→ (A v C)
GOAL: ¬E
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Sign up to join this communityI'm working on an assignment and I'm stuck on this proof. I feel like I'm on the right track but I can't find the way to prove the goal.
B ^ D
(B^¬A) → ¬C
B → ¬A
(D^E)→ (A v C)
GOAL: ¬E
I started by ^elim for B and D, then I did a subproof to start eliminating the conditionals, but how could I prove ~E when ~E is not in the premises? I can't seem to find the logical way to do that.
Good start, but you do not need a subproof to eliminate the conditionals. It is an in-context inference.
Okay, now the goal is ¬E
when that negation may not be directly derived. That is an indication to try an indirect proof (a proof of negation).
So assume E
and seek to derive a contradiction. Let's get you started.
1.| B ^ D premise
2.| (B^¬A) → ¬C premise
3.| B → ¬A premise
4.|_ (D^E) → (A v C) premise
5.| D ^elim 1
6.| B ^elim 2
7.| ¬A →elim 3,5
8.| B ^ ¬A ^intro 6,7
9.| ¬C →elim 2,8
10.| |_ E assumption
11.| | D ^ E ^intro 5,10
12.| | (A v C) →elim 4,11
:| | :
:| | :
| | # ¬elim
| ¬E ¬intro
Well, obviously Proof by Cases is next.
PS: Some Fitch systems do not include a falsum symbol. Their implementation of negation elimination/introduction may differ somewhat, but the basics are the same. If a contradiction is derivable when E
is assumed, then we may infer ¬E
.
A
) because you may reference line 7 (¬A
). A contradiction may also be derived when assuming the right case (C
) because you can reference line 9 (¬C
).
May 13, 2020 at 3:15