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Theorem funsseq 36131
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 3997 . 2 (𝐹 = 𝐺𝐹𝐺)
2 simpl3 1210 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → dom 𝐹 = dom 𝐺)
32reseq2d 5969 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = (𝐺 ↾ dom 𝐺))
4 funssres 6569 . . . . 5 ((Fun 𝐺𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
543ad2antl2 1203 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐹) = 𝐹)
6 simpl2 1209 . . . . 5 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → Fun 𝐺)
7 funrel 6542 . . . . 5 (Fun 𝐺 → Rel 𝐺)
8 resdm 6016 . . . . 5 (Rel 𝐺 → (𝐺 ↾ dom 𝐺) = 𝐺)
96, 7, 83syl 19 . . . 4 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → (𝐺 ↾ dom 𝐺) = 𝐺)
103, 5, 93eqtr3d 2808 . . 3 (((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) ∧ 𝐹𝐺) → 𝐹 = 𝐺)
1110ex 417 . 2 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹𝐺𝐹 = 𝐺))
121, 11impbid2 229 1 ((Fun 𝐹 ∧ Fun 𝐺 ∧ dom 𝐹 = dom 𝐺) → (𝐹 = 𝐺𝐹𝐺))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 209  wa 400  w3a 1101   = wceq 1563  wss 3907  dom cdm 5652  cres 5654  Rel wrel 5657  Fun wfun 6519
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-12 2215  ax-ext 2737  ax-sep 5251  ax-pr 5395
This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1566  df-fal 1576  df-ex 1803  df-sb 2094  df-mo 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-ral 3080  df-rex 3090  df-rab 3418  df-v 3459  df-dif 3910  df-un 3912  df-in 3914  df-ss 3924  df-nul 4289  df-if 4484  df-sn 4586  df-pr 4588  df-op 4592  df-br 5106  df-opab 5168  df-id 5547  df-xp 5658  df-rel 5659  df-cnv 5660  df-co 5661  df-dm 5662  df-res 5664  df-fun 6527
This theorem is referenced by: (None)
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