Theorem funeqd 4923

Description: Equality deduction for the function predicate. (Contributed by NM, 23-Feb-2013.)

Hypothesis

Ref	Expression
funeqd.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
funeqd	⊢ (𝜑 → (Fun 𝐴 ↔ Fun 𝐵))

Proof of Theorem funeqd

Step	Hyp	Ref	Expression
1		funeqd.1	. 2 ⊢ (𝜑 → 𝐴 = 𝐵)
2		funeq 4921	. 2 ⊢ (𝐴 = 𝐵 → (Fun 𝐴 ↔ Fun 𝐵))
3	1, 2	syl 14	1 ⊢ (𝜑 → (Fun 𝐴 ↔ Fun 𝐵))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243  Fun wfun 4896

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-in 2924  df-ss 2931  df-br 3765  df-opab 3819  df-rel 4352  df-cnv 4353  df-co 4354  df-fun 4904

This theorem is referenced by:  funopg  4934  funsng  4946  funcnvuni  4968  f1eq1  5087  shftfn  9425