Theorem eqneqall 2385

Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.)

Assertion

Ref	Expression
eqneqall	⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑))

Proof of Theorem eqneqall

Step	Hyp	Ref	Expression
1		df-ne 2376	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		pm2.24 622	. 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵 → 𝜑))
3	1, 2	biimtrid 152	1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1372   ≠ wne 2375

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 616

This theorem depends on definitions:  df-bi 117  df-ne 2376

This theorem is referenced by:  eldju2ndl  7156  eldju2ndr  7157  modfzo0difsn  10521  nno  12136  prm2orodd  12367  prm23lt5  12505  dvdsprmpweqnn  12578  logbgcd1irr  15357  gausslemma2dlem0f  15449  gausslemma2dlem0i  15452  2lgs  15499  2lgsoddprm  15508