Theorem 2albiim 1612

Description: Split a biconditional and distribute 2 quantifiers. (Contributed by NM, 3-Feb-2005.)

Assertion

Ref	Expression
2albiim	|- (A.xA.y(ph <-> ps) <-> (A.xA.y(ph -> ps) /\ A.xA.y(ps -> ph)))

Proof of Theorem 2albiim

Step	Hyp	Ref	Expression
1		albiim 1611	. . 3 |- (A.y(ph <-> ps) <-> (A.y(ph -> ps) /\ A.y(ps -> ph)))
2	1	albii 1566	. 2 |- (A.xA.y(ph <-> ps) <-> A.x(A.y(ph -> ps) /\ A.y(ps -> ph)))
3		19.26 1593	. 2 |- (A.x(A.y(ph -> ps) /\ A.y(ps -> ph)) <-> (A.xA.y(ph -> ps) /\ A.xA.y(ps -> ph)))
4	2, 3	bitri 240	1 |- (A.xA.y(ph <-> ps) <-> (A.xA.y(ph -> ps) /\ A.xA.y(ps -> ph)))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358  A.wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557

This theorem depends on definitions:  df-bi 177  df-an 360

This theorem is referenced by:  sbnf2  2108  2eu6  2289  eqrel  4846  eqopr  4848