Theorem sbel2x 1871

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

Assertion

Ref	Expression
sbel2x	⊢ (φ ↔ ∃x∃y((x = z ∧ y = w) ∧ [y / w][x / z]φ))

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

Allowed substitution hints:   φ(z,w)

Proof of Theorem sbel2x

Step	Hyp	Ref	Expression
1		sbelx 1870	. . . . 5 ⊢ ([x / z]φ ↔ ∃y(y = w ∧ [y / w][x / z]φ))
2	1	anbi2i 430	. . . 4 ⊢ ((x = z ∧ [x / z]φ) ↔ (x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
3	2	exbii 1493	. . 3 ⊢ (∃x(x = z ∧ [x / z]φ) ↔ ∃x(x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
4		sbelx 1870	. . 3 ⊢ (φ ↔ ∃x(x = z ∧ [x / z]φ))
5		exdistr 1784	. . 3 ⊢ (∃x∃y(x = z ∧ (y = w ∧ [y / w][x / z]φ)) ↔ ∃x(x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
6	3, 4, 5	3bitr4i 201	. 2 ⊢ (φ ↔ ∃x∃y(x = z ∧ (y = w ∧ [y / w][x / z]φ)))
7		anass 381	. . 3 ⊢ (((x = z ∧ y = w) ∧ [y / w][x / z]φ) ↔ (x = z ∧ (y = w ∧ [y / w][x / z]φ)))
8	7	2exbii 1494	. 2 ⊢ (∃x∃y((x = z ∧ y = w) ∧ [y / w][x / z]φ) ↔ ∃x∃y(x = z ∧ (y = w ∧ [y / w][x / z]φ)))
9	6, 8	bitr4i 176	1 ⊢ (φ ↔ ∃x∃y((x = z ∧ y = w) ∧ [y / w][x / z]φ))

Syntax hints:   ∧ wa 97   ↔ wb 98  ∃wex 1378  [wsb 1642

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424

This theorem depends on definitions:  df-bi 110  df-sb 1643

This theorem is referenced by: (None)