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Theorem feq2 5132
Description: Equality theorem for functions. (Contributed by NM, 1-Aug-1994.)
Assertion
Ref Expression
feq2 (𝐴 = 𝐵 → (𝐹:𝐴𝐶𝐹:𝐵𝐶))

Proof of Theorem feq2
StepHypRef Expression
1 fneq2 5089 . . 3 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
21anbi1d 453 . 2 (𝐴 = 𝐵 → ((𝐹 Fn 𝐴 ∧ ran 𝐹𝐶) ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹𝐶)))
3 df-f 5006 . 2 (𝐹:𝐴𝐶 ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹𝐶))
4 df-f 5006 . 2 (𝐹:𝐵𝐶 ↔ (𝐹 Fn 𝐵 ∧ ran 𝐹𝐶))
52, 3, 43bitr4g 221 1 (𝐴 = 𝐵 → (𝐹:𝐴𝐶𝐹:𝐵𝐶))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103   = wceq 1289  wss 2997  ran crn 4429   Fn wfn 4997  wf 4998
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-4 1445  ax-17 1464  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-cleq 2081  df-fn 5005  df-f 5006
This theorem is referenced by:  feq23  5134  feq2d  5136  feq2i  5141  f00  5186  f0dom0  5188  f1eq2  5196  fressnfv  5468  tfrcllemsucfn  6100  tfrcllemsucaccv  6101  tfrcllembxssdm  6103  tfrcllembfn  6104  tfrcllemaccex  6108  tfrcllemres  6109  tfrcldm  6110  tfrcl  6111  mapvalg  6395  map0g  6425  ac6sfi  6594  isomni  6771
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