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Theorem fneqeql 5218
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 5208 . . 3 ((𝐹 Fn A 𝐺 Fn A) → (𝐹 = 𝐺x A (𝐹x) = (𝐺x)))
2 eqcom 2039 . . . 4 ({x A ∣ (𝐹x) = (𝐺x)} = AA = {x A ∣ (𝐹x) = (𝐺x)})
3 rabid2 2480 . . . 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 5217 . . 3 ((𝐹 Fn A 𝐺 Fn A) → dom (𝐹𝐺) = {x A ∣ (𝐹x) = (𝐺x)})
76eqeq1d 2045 . 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 1242  wral 2300  {crab 2304  cin 2910  dom cdm 4288   Fn wfn 4840  cfv 4845
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 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-14 1402  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019  ax-sep 3866  ax-pow 3918  ax-pr 3935
This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-eu 1900  df-mo 1901  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-rex 2306  df-rab 2309  df-v 2553  df-sbc 2759  df-csb 2847  df-un 2916  df-in 2918  df-ss 2925  df-pw 3353  df-sn 3373  df-pr 3374  df-op 3376  df-uni 3572  df-br 3756  df-opab 3810  df-mpt 3811  df-id 4021  df-xp 4294  df-rel 4295  df-cnv 4296  df-co 4297  df-dm 4298  df-iota 4810  df-fun 4847  df-fn 4848  df-fv 4853
This theorem is referenced by:  fneqeql2  5219  fnreseql  5220
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