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Theorem funsseq 36082
Description: Given two functions with equal domains, equality only requires one direction of the subset relationship. (Contributed by Scott Fenton, 24-Apr-2012.) (Proof shortened by Mario Carneiro, 3-May-2015.)
Assertion
Ref Expression
funsseq ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹 = 𝐺𝐹𝐺))

Proof of Theorem funsseq
StepHypRef Expression
1 eqimss 3994 . 2 (𝐹 = 𝐺𝐹𝐺)
2 simpl3 1206 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → dom 𝐹 = dom 𝐺)
32reseq2d 5963 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = (𝐺 ↾ dom 𝐺))
4 funssres 6561 . . . . 5 ((Fun 𝐺𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
543ad2antl2 1199 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
6 simpl2 1205 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → Fun 𝐺)
7 funrel 6534 . . . . 5 (Fun 𝐺 → Rel 𝐺)
8 resdm 6010 . . . . 5 (Rel 𝐺 → (𝐺 ↾ dom 𝐺) = 𝐺)
96, 7, 83syl 18 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐺) = 𝐺)
103, 5, 93eqtr3d 2804 . . 3 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → 𝐹 = 𝐺)
1110ex 416 . 2 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹𝐺𝐹 = 𝐺))
121, 11impbid2 228 1 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹 = 𝐺𝐹𝐺))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 208  wa 399  w3a 1097   = wceq 1559  wss 3904  dom cdm 5645  cres 5647  Rel wrel 5650  Fun wfun 6511
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-12 2211  ax-ext 2733  ax-sep 5245  ax-pr 5389
This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-br 5100  df-opab 5162  df-id 5540  df-xp 5651  df-rel 5652  df-cnv 5653  df-co 5654  df-dm 5655  df-res 5657  df-fun 6519
This theorem is referenced by: (None)
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