Theorem feq2 2749

Description: Equality theorem for functions.

Assertion

Ref	Expression
feq2	⊢ (A = B → (F:A–→C ↔ F:B–→C))

Proof of Theorem feq2

Step	Hyp	Ref	Expression
1		fneq2 2719	. . 3 ⊢ (A = B → (F Fn A ↔ F Fn B))
2	1	anbi1d 469	. 2 ⊢ (A = B → ((F Fn A ∧ ran F ⊆ C) ↔ (F Fn B ∧ ran F ⊆ C)))
3		df-f 2434	. 2 ⊢ (F:A–→C ↔ (F Fn A ∧ ran F ⊆ C))
4		df-f 2434	. 2 ⊢ (F:B–→C ↔ (F Fn B ∧ ran F ⊆ C))
5	2, 3, 4	3bitr4g 428	1 ⊢ (A = B → (F:A–→C ↔ F:B–→C))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196   = wceq 1091   ⊆ wss 1487  ran crn 2411   Fn wfn 2417  –→wf 2418

This theorem is referenced by:  f00 2773  fconst 2774  f1eq2 2777  fressnfv 2898  fconstfv 2903  mapvalg 3263  mapdom2 3389

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-cleq 1097  df-fn 2433  df-f 2434

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