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Theorem fneqeql 6793
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 6779 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥)))
2 eqcom 2805 . . . 4 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
3 rabid2 3334 . . . 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 6792 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
76eqeq1d 2800 . 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 3106  {crab 3110  cin 3880  dom cdm 5519   Fn wfn 6319  cfv 6324
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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770  ax-sep 5167  ax-nul 5174  ax-pow 5231  ax-pr 5295
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 2070  df-mo 2598  df-eu 2629  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-ral 3111  df-rex 3112  df-rab 3115  df-v 3443  df-sbc 3721  df-csb 3829  df-dif 3884  df-un 3886  df-in 3888  df-ss 3898  df-nul 4244  df-if 4426  df-sn 4526  df-pr 4528  df-op 4532  df-uni 4801  df-br 5031  df-opab 5093  df-mpt 5111  df-id 5425  df-xp 5525  df-rel 5526  df-cnv 5527  df-co 5528  df-dm 5529  df-rn 5530  df-res 5531  df-ima 5532  df-iota 6283  df-fun 6326  df-fn 6327  df-fv 6332
This theorem is referenced by:  fneqeql2  6794  fnreseql  6795  lspextmo  19821
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