Theorem sbel2x 2125

Description: Elimination of double substitution. (Contributed by NM, 5-Aug-1993.)

Assertion

Ref	Expression
sbel2x	|- (ph <-> E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph))

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

Allowed substitution hints:   ph(z,w)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 2124	. . . . 5 |- ([x / z]ph <-> E.y(y = w /\ [y / w][x / z]ph))
2	1	anbi2i 675	. . . 4 |- ((x = z /\ [x / z]ph) <-> (x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
3	2	exbii 1582	. . 3 |- (E.x(x = z /\ [x / z]ph) <-> E.x(x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
4		sbelx 2124	. . 3 |- (ph <-> E.x(x = z /\ [x / z]ph))
5		exdistr 1906	. . 3 |- (E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)) <-> E.x(x = z /\ E.y(y = w /\ [y / w][x / z]ph)))
6	3, 4, 5	3bitr4i 268	. 2 |- (ph <-> E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)))
7		anass 630	. . 3 |- (((x = z /\ y = w) /\ [y / w][x / z]ph) <-> (x = z /\ (y = w /\ [y / w][x / z]ph)))
8	7	2exbii 1583	. 2 |- (E.xE.y((x = z /\ y = w) /\ [y / w][x / z]ph) <-> E.xE.y(x = z /\ (y = w /\ [y / w][x / z]ph)))
9	6, 8	bitr4i 243	1 |- (ph <-> E.xE.y((x = z /\ y = w) /\ [y / w][x / z]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: (None)