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Theorem funsseq 36003
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 3980 . 2 (𝐹 = 𝐺𝐹𝐺)
2 simpl3 1200 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → dom 𝐹 = dom 𝐺)
32reseq2d 5938 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = (𝐺 ↾ dom 𝐺))
4 funssres 6536 . . . . 5 ((Fun 𝐺𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
543ad2antl2 1193 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
6 simpl2 1199 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → Fun 𝐺)
7 funrel 6509 . . . . 5 (Fun 𝐺 → Rel 𝐺)
8 resdm 5985 . . . . 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 2783 . . 3 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → 𝐹 = 𝐺)
1110ex 413 . 2 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹𝐺𝐹 = 𝐺))
121, 11impbid2 227 1 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹 = 𝐺𝐹𝐺))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 207  wa 396  w3a 1092   = wceq 1547  wss 3890  dom cdm 5625  cres 5627  Rel wrel 5630  Fun wfun 6486
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1974  ax-7 2015  ax-8 2121  ax-9 2129  ax-12 2189  ax-ext 2712  ax-sep 5225  ax-pr 5369
This theorem depends on definitions:  df-bi 208  df-an 397  df-or 854  df-3an 1094  df-tru 1550  df-fal 1560  df-ex 1787  df-sb 2074  df-mo 2543  df-eu 2573  df-clab 2719  df-cleq 2732  df-clel 2815  df-ral 3055  df-rex 3065  df-rab 3393  df-v 3434  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4269  df-if 4462  df-sn 4563  df-pr 4565  df-op 4569  df-br 5080  df-opab 5142  df-id 5520  df-xp 5631  df-rel 5632  df-cnv 5633  df-co 5634  df-dm 5635  df-res 5637  df-fun 6494
This theorem is referenced by: (None)
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