Theorem 2sb5 2112

Description: Equivalence for double substitution. (Contributed by NM, 3-Feb-2005.)

Assertion

Ref	Expression
2sb5	|- ([z / x][w / y]ph <-> E.xE.y((x = z /\ y = w) /\ ph))

Distinct variable groups:   x,y,z   y,w

Allowed substitution hints:   ph(x,y,z,w)

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 2100	. 2 |- ([z / x][w / y]ph <-> E.x(x = z /\ [w / y]ph))
2		19.42v 1905	. . . 4 |- (E.y(x = z /\ (y = w /\ ph)) <-> (x = z /\ E.y(y = w /\ ph)))
3		anass 630	. . . . 5 |- (((x = z /\ y = w) /\ ph) <-> (x = z /\ (y = w /\ ph)))
4	3	exbii 1582	. . . 4 |- (E.y((x = z /\ y = w) /\ ph) <-> E.y(x = z /\ (y = w /\ ph)))
5		sb5 2100	. . . . 5 |- ([w / y]ph <-> E.y(y = w /\ ph))
6	5	anbi2i 675	. . . 4 |- ((x = z /\ [w / y]ph) <-> (x = z /\ E.y(y = w /\ ph)))
7	2, 4, 6	3bitr4ri 269	. . 3 |- ((x = z /\ [w / y]ph) <-> E.y((x = z /\ y = w) /\ ph))
8	7	exbii 1582	. 2 |- (E.x(x = z /\ [w / y]ph) <-> E.xE.y((x = z /\ y = w) /\ ph))
9	1, 8	bitri 240	1 |- ([z / x][w / y]ph <-> E.xE.y((x = z /\ y = w) /\ ph))

Syntax hints:   <-> wb 176   /\ wa 358  E.wex 1541  [wsb 1648

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

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  pm11.07  2115