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Theorem fneq2d 5289
Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)
Hypothesis
Ref Expression
fneq2d.1 (𝜑𝐴 = 𝐵)
Assertion
Ref Expression
fneq2d (𝜑 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))

Proof of Theorem fneq2d
StepHypRef Expression
1 fneq2d.1 . 2 (𝜑𝐴 = 𝐵)
2 fneq2 5287 . 2 (𝐴 = 𝐵 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
31, 2syl 14 1 (𝜑 → (𝐹 Fn 𝐴𝐹 Fn 𝐵))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104   = wceq 1348   Fn wfn 5193
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-gen 1442  ax-4 1503  ax-17 1519  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-cleq 2163  df-fn 5201
This theorem is referenced by:  fneq12d  5290  acfun  7184  ccfunen  7226  seq3shft  10802
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