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Theorem fneqeql 6807
Description: Two functions are equal iff their equalizer is the whole domain. (Contributed by Stefan O'Rear, 7-Mar-2015.)
Assertion
Ref Expression
fneqeql ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))

Proof of Theorem fneqeql
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 eqfnfv 6793 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥)))
2 eqcom 2831 . . . 4 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
3 rabid2 3372 . . . 4 (𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
42, 3bitri 278 . . 3 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
51, 4syl6bbr 292 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
6 fndmin 6806 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
76eqeq1d 2826 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (dom (𝐹𝐺) = 𝐴 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
85, 7bitr4d 285 1 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 399   = wceq 1538  wral 3133  {crab 3137  cin 3918  dom cdm 5542   Fn wfn 6338  cfv 6343
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2179  ax-ext 2796  ax-sep 5189  ax-nul 5196  ax-pow 5253  ax-pr 5317
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-mo 2624  df-eu 2655  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2964  df-ral 3138  df-rex 3139  df-rab 3142  df-v 3482  df-sbc 3759  df-csb 3867  df-dif 3922  df-un 3924  df-in 3926  df-ss 3936  df-nul 4277  df-if 4451  df-sn 4551  df-pr 4553  df-op 4557  df-uni 4825  df-br 5053  df-opab 5115  df-mpt 5133  df-id 5447  df-xp 5548  df-rel 5549  df-cnv 5550  df-co 5551  df-dm 5552  df-rn 5553  df-res 5554  df-ima 5555  df-iota 6302  df-fun 6345  df-fn 6346  df-fv 6351
This theorem is referenced by:  fneqeql2  6808  fnreseql  6809  lspextmo  19828
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