Theorem eqneqall 2259

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 2250	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵)
2		pm2.24 584	. 2 ⊢ (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵 → 𝜑))
3	1, 2	syl5bi 150	1 ⊢ (𝐴 = 𝐵 → (𝐴 ≠ 𝐵 → 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   = wceq 1285   ≠ wne 2249

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-in2 578

This theorem depends on definitions:  df-bi 115  df-ne 2250

This theorem is referenced by:  eldju2ndl  6566  eldju2ndr  6567  modfzo0difsn  9547  nno  10531  prm2orodd  10733