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Theorem eqneqall 2370
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 2361 . 2 (𝐴𝐵 ↔ ¬ 𝐴 = 𝐵)
2 pm2.24 622 . 2 (𝐴 = 𝐵 → (¬ 𝐴 = 𝐵𝜑))
31, 2biimtrid 152 1 (𝐴 = 𝐵 → (𝐴𝐵𝜑))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   = wceq 1364  wne 2360
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 2361
This theorem is referenced by:  eldju2ndl  7117  eldju2ndr  7118  modfzo0difsn  10452  nno  12021  prm2orodd  12238  prm23lt5  12375  dvdsprmpweqnn  12448  logbgcd1irr  15027  gausslemma2dlem0f  15098  gausslemma2dlem0i  15101
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