fneq2 - New Foundations Explorer

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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 A ↔ F Fn B))

**Proof of Theorem fneq2**
Step	Hyp	Ref	Expression
1		eqeq2 2362	. . 3 ⊢ (A = B → (dom F = A ↔ dom F = B))
2	1	anbi2d 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))
5	2, 3, 4	3bitr4g 279	1 ⊢ (A = B → (F Fn A ↔ F 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-1 5 ax-2 6 ax-3 7 ax-mp 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