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Theorem fneqeql 6808
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 6794 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥)))
2 eqcom 2825 . . . 4 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
3 rabid2 3379 . . . 4 (𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
42, 3bitri 276 . . 3 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
51, 4syl6bbr 290 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
6 fndmin 6807 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
76eqeq1d 2820 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (dom (𝐹𝐺) = 𝐴 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
85, 7bitr4d 283 1 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396   = wceq 1528  wral 3135  {crab 3139  cin 3932  dom cdm 5548   Fn wfn 6343  cfv 6348
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1787  ax-4 1801  ax-5 1902  ax-6 1961  ax-7 2006  ax-8 2107  ax-9 2115  ax-10 2136  ax-11 2151  ax-12 2167  ax-ext 2790  ax-sep 5194  ax-nul 5201  ax-pow 5257  ax-pr 5320
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 842  df-3an 1081  df-tru 1531  df-ex 1772  df-nf 1776  df-sb 2061  df-mo 2615  df-eu 2647  df-clab 2797  df-cleq 2811  df-clel 2890  df-nfc 2960  df-ral 3140  df-rex 3141  df-rab 3144  df-v 3494  df-sbc 3770  df-csb 3881  df-dif 3936  df-un 3938  df-in 3940  df-ss 3949  df-nul 4289  df-if 4464  df-sn 4558  df-pr 4560  df-op 4564  df-uni 4831  df-br 5058  df-opab 5120  df-mpt 5138  df-id 5453  df-xp 5554  df-rel 5555  df-cnv 5556  df-co 5557  df-dm 5558  df-rn 5559  df-res 5560  df-ima 5561  df-iota 6307  df-fun 6350  df-fn 6351  df-fv 6356
This theorem is referenced by:  fneqeql2  6809  fnreseql  6810  lspextmo  19757
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