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Theorem fneq2 5019
Description: Equality theorem for function predicate with domain. (Contributed by NM, 1-Aug-1994.)
Assertion
Ref Expression
fneq2 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))

Proof of Theorem fneq2
StepHypRef Expression
1 eqeq2 2091 . . 3 (𝐴 = 𝐵 → (dom 𝐹 = 𝐴 ↔ dom 𝐹 = 𝐵))
21anbi2d 452 . 2 (𝐴 = 𝐵 → ((Fun 𝐹 ∧ dom 𝐹 = 𝐴) ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵)))
3 df-fn 4935 . 2 (𝐹 Fn 𝐴 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐴))
4 df-fn 4935 . 2 (𝐹 Fn 𝐵 ↔ (Fun 𝐹 ∧ dom 𝐹 = 𝐵))
52, 3, 43bitr4g 221 1 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103   = wceq 1285  dom cdm 4371  Fun wfun 4926   Fn wfn 4927
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 1377  ax-gen 1379  ax-4 1441  ax-17 1460  ax-ext 2064
This theorem depends on definitions:  df-bi 115  df-cleq 2075  df-fn 4935
This theorem is referenced by:  fneq2d  5021  fneq2i  5025  feq2  5062  foeq2  5134  f1o00  5192  eqfnfv2  5298  tfr0dm  5971  tfrlemisucaccv  5974  tfrlemi1  5981  tfrlemi14d  5982  tfrexlem  5983  tfr1onlemsucfn  5989  tfr1onlemsucaccv  5990  tfr1onlembxssdm  5992  tfr1onlembfn  5993  tfr1onlemaccex  5997  tfr1onlemres  5998  0fz1  9140
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