Theorem ifcl 3699

Description: Membership (closure) of a conditional operator. (Contributed by NM, 4-Apr-2005.)

Assertion

Ref	Expression
ifcl	|- ((A e. C /\ B e. C) -> if(ph, A, B) e. C)

Proof of Theorem ifcl

Step	Hyp	Ref	Expression
1		eleq1 2413	. 2 |- (A = if(ph, A, B) -> (A e. C <-> if(ph, A, B) e. C))
2		eleq1 2413	. 2 |- (B = if(ph, A, B) -> (B e. C <-> if(ph, A, B) e. C))
3	1, 2	ifboth 3694	1 |- ((A e. C /\ B e. C) -> if(ph, A, B) e. C)

Syntax hints:   -> wi 4   /\ wa 358   e. wcel 1710  ifcif 3663

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-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-if 3664

This theorem is referenced by:  ifpr  3775