Theorem sbel2x 2125

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 2124	. . . . 5 ⊢ ([x / z]φ ↔ ∃y(y = w ∧ [y / w][x / z]φ))
2	1	anbi2i 675	. . . 4 ⊢ ((x = z ∧ [x / z]φ) ↔ (x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
3	2	exbii 1582	. . 3 ⊢ (∃x(x = z ∧ [x / z]φ) ↔ ∃x(x = z ∧ ∃y(y = w ∧ [y / w][x / z]φ)))
4		sbelx 2124	. . 3 ⊢ (φ ↔ ∃x(x = z ∧ [x / z]φ))
5		exdistr 1906	. . 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 268	. 2 ⊢ (φ ↔ ∃x∃y(x = z ∧ (y = w ∧ [y / w][x / z]φ)))
7		anass 630	. . 3 ⊢ (((x = z ∧ y = w) ∧ [y / w][x / z]φ) ↔ (x = z ∧ (y = w ∧ [y / w][x / z]φ)))
8	7	2exbii 1583	. 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 243	1 ⊢ (φ ↔ ∃x∃y((x = z ∧ y = w) ∧ [y / w][x / z]φ))

This proof depends on syntax axioms:   ↔ wb 176   ∧ wa 358  ∃wex 1541  [wsb 1648

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

This theorem is used by: (None)