Theorem eqneqall 2410

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 2401	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		pm2.24 624	. 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵 → 𝜑))
3	1, 2	biimtrid 152	1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1395   ≠ wne 2400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-in2 618

This theorem depends on definitions:  df-bi 117  df-ne 2401

This theorem is referenced by:  ssprsseq  3829  eldju2ndl  7235  eldju2ndr  7236  modfzo0difsn  10612  nno  12412  prm2orodd  12643  prm23lt5  12781  dvdsprmpweqnn  12854  logbgcd1irr  15635  gausslemma2dlem0f  15727  gausslemma2dlem0i  15730  2lgs  15777  2lgsoddprm  15786  umgrnloop2  15943  uhgr2edg  15998