Theorem fneq2d 5472

Description: Equality deduction for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq2d.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
fneq2d	⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

Proof of Theorem fneq2d

Step	Hyp	Ref	Expression
1		fneq2d.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		fneq2 5470	. 2 ⊢ (𝐴 = 𝐵 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (𝐹 Fn 𝐴 ↔ 𝐹 Fn 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 105   = wceq 1402   Fn wfn 5372

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-4 1563  ax-17 1579  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-cleq 2231  df-fn 5380

This theorem is used by:  fneq12d  5473  fncofn  5893  acfun  7563  ccfunen  7630  ccatlid  11376  ccatrid  11377  ccatass  11378  ccatswrd  11444  swrdccat2  11445  ccatpfx  11475  swrdswrd  11479  swrdccatin2  11503  pfxccatin12  11507  seq3shft  11605  ptex  13620  rng1zrlem  14260