Theorem 2sb5 1856

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

Assertion

Ref	Expression
2sb5	⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ∧ y = w) ∧ φ))

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

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

Proof of Theorem 2sb5

Step	Hyp	Ref	Expression
1		sb5 1764	. 2 ⊢ ([z / x][w / y]φ ↔ ∃x(x = z ∧ [w / y]φ))
2		19.42v 1783	. . . 4 ⊢ (∃y(x = z ∧ (y = w ∧ φ)) ↔ (x = z ∧ ∃y(y = w ∧ φ)))
3		anass 381	. . . . 5 ⊢ (((x = z ∧ y = w) ∧ φ) ↔ (x = z ∧ (y = w ∧ φ)))
4	3	exbii 1493	. . . 4 ⊢ (∃y((x = z ∧ y = w) ∧ φ) ↔ ∃y(x = z ∧ (y = w ∧ φ)))
5		sb5 1764	. . . . 5 ⊢ ([w / y]φ ↔ ∃y(y = w ∧ φ))
6	5	anbi2i 430	. . . 4 ⊢ ((x = z ∧ [w / y]φ) ↔ (x = z ∧ ∃y(y = w ∧ φ)))
7	2, 4, 6	3bitr4ri 202	. . 3 ⊢ ((x = z ∧ [w / y]φ) ↔ ∃y((x = z ∧ y = w) ∧ φ))
8	7	exbii 1493	. 2 ⊢ (∃x(x = z ∧ [w / y]φ) ↔ ∃x∃y((x = z ∧ y = w) ∧ φ))
9	1, 8	bitri 173	1 ⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ∧ y = w) ∧ 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:  opelopabsbALT  3987