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Theorem fneqeql 6588
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 6574 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥)))
2 eqcom 2785 . . . 4 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
3 rabid2 3305 . . . 4 (𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
42, 3bitri 267 . . 3 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
51, 4syl6bbr 281 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
6 fndmin 6587 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
76eqeq1d 2780 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (dom (𝐹𝐺) = 𝐴 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
85, 7bitr4d 274 1 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wa 386   = wceq 1601  wral 3090  {crab 3094  cin 3791  dom cdm 5355   Fn wfn 6130  cfv 6135
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1839  ax-4 1853  ax-5 1953  ax-6 2021  ax-7 2055  ax-8 2109  ax-9 2116  ax-10 2135  ax-11 2150  ax-12 2163  ax-13 2334  ax-ext 2754  ax-sep 5017  ax-nul 5025  ax-pow 5077  ax-pr 5138
This theorem depends on definitions:  df-bi 199  df-an 387  df-or 837  df-3an 1073  df-tru 1605  df-ex 1824  df-nf 1828  df-sb 2012  df-mo 2551  df-eu 2587  df-clab 2764  df-cleq 2770  df-clel 2774  df-nfc 2921  df-ral 3095  df-rex 3096  df-rab 3099  df-v 3400  df-sbc 3653  df-csb 3752  df-dif 3795  df-un 3797  df-in 3799  df-ss 3806  df-nul 4142  df-if 4308  df-sn 4399  df-pr 4401  df-op 4405  df-uni 4672  df-br 4887  df-opab 4949  df-mpt 4966  df-id 5261  df-xp 5361  df-rel 5362  df-cnv 5363  df-co 5364  df-dm 5365  df-rn 5366  df-res 5367  df-ima 5368  df-iota 6099  df-fun 6137  df-fn 6138  df-fv 6143
This theorem is referenced by:  fneqeql2  6589  fnreseql  6590  lspextmo  19451
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