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Theorem fneqeql 7042
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 7026 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥)))
2 eqcom 2769 . . . 4 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
3 rabid2 3447 . . . 4 (𝐴 = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
42, 3bitri 278 . . 3 ({𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴 ↔ ∀𝑥𝐴 (𝐹𝑥) = (𝐺𝑥))
51, 4bitr4di 292 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)} = 𝐴))
6 fndmin 7041 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) = {𝑥𝐴 ∣ (𝐹𝑥) = (𝐺𝑥)})
76eqeq1d 2764 . 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
This proof depends on syntax axioms:  wi 4  wb 209  wa 401   = wceq 1570  wral 3078  {crab 3414  cin 3901  dom cdm 5659   Fn wfn 6532  cfv 6537
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2734  ax-sep 5255  ax-nul 5267  ax-pr 5402
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-sbc 3743  df-csb 3851  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-mpt 5191  df-id 5554  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-res 5671  df-ima 5672  df-iota 6493  df-fun 6539  df-fn 6540  df-fv 6545
This theorem is used by:  fneqeql2  7043  fnreseql  7044  lspextmo  21246
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