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Theorem fneqeql 5196
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 A 𝐺 Fn A) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = A))

Proof of Theorem fneqeql
Dummy variable x is distinct from all other variables.
StepHypRef Expression
1 eqfnfv 5186 . . 3 ((𝐹 Fn A 𝐺 Fn A) → (𝐹 = 𝐺x A (𝐹x) = (𝐺x)))
2 eqcom 2020 . . . 4 ({x A ∣ (𝐹x) = (𝐺x)} = AA = {x A ∣ (𝐹x) = (𝐺x)})
3 rabid2 2460 . . . 4 (A = {x A ∣ (𝐹x) = (𝐺x)} ↔ x A (𝐹x) = (𝐺x))
42, 3bitri 173 . . 3 ({x A ∣ (𝐹x) = (𝐺x)} = Ax A (𝐹x) = (𝐺x))
51, 4syl6bbr 187 . 2 ((𝐹 Fn A 𝐺 Fn A) → (𝐹 = 𝐺 ↔ {x A ∣ (𝐹x) = (𝐺x)} = A))
6 fndmin 5195 . . 3 ((𝐹 Fn A 𝐺 Fn A) → dom (𝐹𝐺) = {x A ∣ (𝐹x) = (𝐺x)})
76eqeq1d 2026 . 2 ((𝐹 Fn A 𝐺 Fn A) → (dom (𝐹𝐺) = A ↔ {x A ∣ (𝐹x) = (𝐺x)} = A))
85, 7bitr4d 180 1 ((𝐹 Fn A 𝐺 Fn A) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = A))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98   = wceq 1226  wral 2280  {crab 2284  cin 2889  dom cdm 4268   Fn wfn 4820  cfv 4825
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 617  ax-5 1312  ax-7 1313  ax-gen 1314  ax-ie1 1359  ax-ie2 1360  ax-8 1372  ax-10 1373  ax-11 1374  ax-i12 1375  ax-bnd 1376  ax-4 1377  ax-14 1382  ax-17 1396  ax-i9 1400  ax-ial 1405  ax-i5r 1406  ax-ext 2000  ax-sep 3845  ax-pow 3897  ax-pr 3914
This theorem depends on definitions:  df-bi 110  df-3an 873  df-tru 1229  df-nf 1326  df-sb 1624  df-eu 1881  df-mo 1882  df-clab 2005  df-cleq 2011  df-clel 2014  df-nfc 2145  df-ral 2285  df-rex 2286  df-rab 2289  df-v 2533  df-sbc 2738  df-csb 2826  df-un 2895  df-in 2897  df-ss 2904  df-pw 3332  df-sn 3352  df-pr 3353  df-op 3355  df-uni 3551  df-br 3735  df-opab 3789  df-mpt 3790  df-id 4000  df-xp 4274  df-rel 4275  df-cnv 4276  df-co 4277  df-dm 4278  df-iota 4790  df-fun 4827  df-fn 4828  df-fv 4833
This theorem is referenced by:  fneqeql2  5197  fnreseql  5198
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