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Theorem funsseq 25394
 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

Proof of Theorem funsseq
StepHypRef Expression
1 eqimss 3401 . 2
2 simpl3 963 . . . . 5
32reseq2d 5147 . . . 4
4 funssres 5494 . . . . 5
543ad2antl2 1121 . . . 4
6 simpl2 962 . . . . 5
7 funrel 5472 . . . . 5
8 resdm 5185 . . . . 5
96, 7, 83syl 19 . . . 4
103, 5, 93eqtr3d 2477 . . 3
1110ex 425 . 2
121, 11impbid2 197 1
 Colors of variables: wff set class Syntax hints:   wi 4   wb 178   wa 360   w3a 937   wceq 1653   wss 3321   cdm 4879   cres 4881   wrel 4884   wfun 5449 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1556  ax-5 1567  ax-17 1627  ax-9 1667  ax-8 1688  ax-14 1730  ax-6 1745  ax-7 1750  ax-11 1762  ax-12 1951  ax-ext 2418  ax-sep 4331  ax-nul 4339  ax-pr 4404 This theorem depends on definitions:  df-bi 179  df-or 361  df-an 362  df-3an 939  df-tru 1329  df-ex 1552  df-nf 1555  df-sb 1660  df-eu 2286  df-mo 2287  df-clab 2424  df-cleq 2430  df-clel 2433  df-nfc 2562  df-ne 2602  df-ral 2711  df-rex 2712  df-rab 2715  df-v 2959  df-dif 3324  df-un 3326  df-in 3328  df-ss 3335  df-nul 3630  df-if 3741  df-sn 3821  df-pr 3822  df-op 3824  df-br 4214  df-opab 4268  df-id 4499  df-xp 4885  df-rel 4886  df-cnv 4887  df-co 4888  df-dm 4889  df-res 4891  df-fun 5457
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