Theorem funeqd 5130

Description: Equality deduction for the function predicate. (Contributed by set.mm contributors, 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 5128	. 2 ⊢ (A = B → (Fun A ↔ Fun B))
3	1, 2	syl 15	1 ⊢ (φ → (Fun A ↔ Fun B))

Syntax hints:   → wi 4   ↔ wb 176   = wceq 1642  Fun wfun 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-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260  df-opab 4624  df-br 4641  df-co 4727  df-cnv 4786  df-fun 4790

This theorem is referenced by:  funcnvuni  5162  f1eq1  5254  fvopab4t  5386