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Theorem fneq2 5174
 Description: Equality theorem for function predicate with domain. (Contributed by set.mm contributors, 1-Aug-1994.)
Assertion
Ref Expression
fneq2 (A = B → (F Fn AF Fn B))

Proof of Theorem fneq2
StepHypRef Expression
1 eqeq2 2362 . . 3 (A = B → (dom F = A ↔ dom F = B))
21anbi2d 684 . 2 (A = B → ((Fun F dom F = A) ↔ (Fun F dom F = B)))
3 df-fn 4790 . 2 (F Fn A ↔ (Fun F dom F = A))
4 df-fn 4790 . 2 (F Fn B ↔ (Fun F dom F = B))
52, 3, 43bitr4g 279 1 (A = B → (F Fn AF Fn B))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358   = wceq 1642  dom cdm 4772  Fun wfun 4775   Fn wfn 4776 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-cleq 2346  df-fn 4790 This theorem is referenced by:  fneq2d  5176  fneq2i  5179  feq2  5211  foeq2  5266  f1o00  5317  eqfnfv2  5393  fconstfv  5456  composefn  5818  brfns  5833  fnfullfun  5858
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