Metamath Proof Explorer


Theorem necon3bid

Description: Deduction from equality to inequality. (Contributed by NM, 23-Feb-2005) (Proof shortened by Andrew Salmon, 25-May-2011)

Ref Expression
Hypothesis necon3bid.1 φ A = B C = D
Assertion necon3bid φ A B C D

Proof

Step Hyp Ref Expression
1 necon3bid.1 φ A = B C = D
2 df-ne A B ¬ A = B
3 1 necon3bbid φ ¬ A = B C D
4 2 3 syl5bb φ A B C D