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Theorem fneqeql2 6819
Description: Two functions are equal iff their equalizer contains the whole domain. (Contributed by Stefan O'Rear, 9-Mar-2015.)
Assertion
Ref Expression
fneqeql2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺𝐴 ⊆ dom (𝐹𝐺)))

Proof of Theorem fneqeql2
StepHypRef Expression
1 fneqeql 6818 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))
2 inss1 4207 . . . . . 6 (𝐹𝐺) ⊆ 𝐹
3 dmss 5773 . . . . . 6 ((𝐹𝐺) ⊆ 𝐹 → dom (𝐹𝐺) ⊆ dom 𝐹)
42, 3ax-mp 5 . . . . 5 dom (𝐹𝐺) ⊆ dom 𝐹
5 fndm 6457 . . . . . 6 (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
65adantr 483 . . . . 5 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom 𝐹 = 𝐴)
74, 6sseqtrid 4021 . . . 4 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) ⊆ 𝐴)
87biantrurd 535 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐴 ⊆ dom (𝐹𝐺) ↔ (dom (𝐹𝐺) ⊆ 𝐴𝐴 ⊆ dom (𝐹𝐺))))
9 eqss 3984 . . 3 (dom (𝐹𝐺) = 𝐴 ↔ (dom (𝐹𝐺) ⊆ 𝐴𝐴 ⊆ dom (𝐹𝐺)))
108, 9syl6rbbr 292 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (dom (𝐹𝐺) = 𝐴𝐴 ⊆ dom (𝐹𝐺)))
111, 10bitrd 281 1 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺𝐴 ⊆ dom (𝐹𝐺)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 398   = wceq 1537  cin 3937  wss 3938  dom cdm 5557   Fn wfn 6352
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795  ax-sep 5205  ax-nul 5212  ax-pow 5268  ax-pr 5332
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-mo 2622  df-eu 2654  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-rex 3146  df-rab 3149  df-v 3498  df-sbc 3775  df-csb 3886  df-dif 3941  df-un 3943  df-in 3945  df-ss 3954  df-nul 4294  df-if 4470  df-sn 4570  df-pr 4572  df-op 4576  df-uni 4841  df-br 5069  df-opab 5131  df-mpt 5149  df-id 5462  df-xp 5563  df-rel 5564  df-cnv 5565  df-co 5566  df-dm 5567  df-rn 5568  df-res 5569  df-ima 5570  df-iota 6316  df-fun 6359  df-fn 6360  df-fv 6365
This theorem is referenced by:  evlseu  20298  hauseqcn  31140
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