Theorem eqneqall 2412

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 2403	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		pm2.24 626	. 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵 → 𝜑))
3	1, 2	biimtrid 152	1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1397   ≠ wne 2402

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 620

This theorem depends on definitions:  df-bi 117  df-ne 2403

This theorem is referenced by:  ssprsseq  3835  eldju2ndl  7270  eldju2ndr  7271  modfzo0difsn  10656  nno  12466  prm2orodd  12697  prm23lt5  12835  dvdsprmpweqnn  12908  logbgcd1irr  15690  gausslemma2dlem0f  15782  gausslemma2dlem0i  15785  2lgs  15832  2lgsoddprm  15841  umgrnloop2  16001  uhgr2edg  16056  umgrclwwlkge2  16252