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Theorem eqneqall 2265
Description: A contradiction concerning equality implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.)
Assertion
Ref Expression
eqneqall (𝐴 = 𝐵 → (𝐴𝐵𝜑))

Proof of Theorem eqneqall
StepHypRef Expression
1 df-ne 2256 . 2 (𝐴𝐵 ↔ ¬ 𝐴 = 𝐵)
2 pm2.24 586 . 2 (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵𝜑))
31, 2syl5bi 150 1 (𝐴 = 𝐵 → (𝐴𝐵𝜑))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1289  wne 2255
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-in2 580
This theorem depends on definitions:  df-bi 115  df-ne 2256
This theorem is referenced by:  eldju2ndl  6742  eldju2ndr  6743  modfzo0difsn  9767  nno  10999  prm2orodd  11201
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