Theorem 2albi 1106

Description: Split a biconditional and distribute 2 quantifiers.

Assertion

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

Proof of Theorem 2albi

Step	Hyp	Ref	Expression
1		albi 1105	. . 3 |- (A.y(ph <-> ps) <-> (A.y(ph -> ps) /\ A.y(ps -> ph)))
2	1	albii 997	. 2 |- (A.xA.y(ph <-> ps) <-> A.x(A.y(ph -> ps) /\ A.y(ps -> ph)))
3		19.26 1065	. 2 |- (A.x(A.y(ph -> ps) /\ A.y(ps -> ph)) <-> (A.xA.y(ph -> ps) /\ A.xA.y(ps -> ph)))
4	2, 3	bitr 173	1 |- (A.xA.y(ph <-> ps) <-> (A.xA.y(ph -> ps) /\ A.xA.y(ps -> ph)))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  A.wal 952

This theorem is referenced by:  2eu6 1452  eqrel 3245

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 961  ax-4 971  ax-5o 973

This theorem depends on definitions:  df-bi 147  df-an 225

Copyright terms: Public domain