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Theorem fneqeql2 6924
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 6923 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹𝐺) = 𝐴))
2 eqss 3936 . . 3 (dom (𝐹𝐺) = 𝐴 ↔ (dom (𝐹𝐺) ⊆ 𝐴𝐴 ⊆ dom (𝐹𝐺)))
3 inss1 4162 . . . . . 6 (𝐹𝐺) ⊆ 𝐹
4 dmss 5811 . . . . . 6 ((𝐹𝐺) ⊆ 𝐹 → dom (𝐹𝐺) ⊆ dom 𝐹)
53, 4ax-mp 5 . . . . 5 dom (𝐹𝐺) ⊆ dom 𝐹
6 fndm 6536 . . . . . 6 (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
76adantr 481 . . . . 5 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom 𝐹 = 𝐴)
85, 7sseqtrid 3973 . . . 4 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → dom (𝐹𝐺) ⊆ 𝐴)
98biantrurd 533 . . 3 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐴 ⊆ dom (𝐹𝐺) ↔ (dom (𝐹𝐺) ⊆ 𝐴𝐴 ⊆ dom (𝐹𝐺))))
102, 9bitr4id 290 . 2 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (dom (𝐹𝐺) = 𝐴𝐴 ⊆ dom (𝐹𝐺)))
111, 10bitrd 278 1 ((𝐹 Fn 𝐴𝐺 Fn 𝐴) → (𝐹 = 𝐺𝐴 ⊆ dom (𝐹𝐺)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396   = wceq 1539  cin 3886  wss 3887  dom cdm 5589   Fn wfn 6428
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2816  df-nfc 2889  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-sbc 3717  df-csb 3833  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-uni 4840  df-br 5075  df-opab 5137  df-mpt 5158  df-id 5489  df-xp 5595  df-rel 5596  df-cnv 5597  df-co 5598  df-dm 5599  df-rn 5600  df-res 5601  df-ima 5602  df-iota 6391  df-fun 6435  df-fn 6436  df-fv 6441
This theorem is referenced by:  evlseu  21293  hauseqcn  31848
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