Theorem 2sb5 2112

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 2100	. 2 ⊢ ([z / x][w / y]φ ↔ ∃x(x = z ∧ [w / y]φ))
2		19.42v 1905	. . . 4 ⊢ (∃y(x = z ∧ (y = w ∧ φ)) ↔ (x = z ∧ ∃y(y = w ∧ φ)))
3		anass 630	. . . . 5 ⊢ (((x = z ∧ y = w) ∧ φ) ↔ (x = z ∧ (y = w ∧ φ)))
4	3	exbii 1582	. . . 4 ⊢ (∃y((x = z ∧ y = w) ∧ φ) ↔ ∃y(x = z ∧ (y = w ∧ φ)))
5		sb5 2100	. . . . 5 ⊢ ([w / y]φ ↔ ∃y(y = w ∧ φ))
6	5	anbi2i 675	. . . 4 ⊢ ((x = z ∧ [w / y]φ) ↔ (x = z ∧ ∃y(y = w ∧ φ)))
7	2, 4, 6	3bitr4ri 269	. . 3 ⊢ ((x = z ∧ [w / y]φ) ↔ ∃y((x = z ∧ y = w) ∧ φ))
8	7	exbii 1582	. 2 ⊢ (∃x(x = z ∧ [w / y]φ) ↔ ∃x∃y((x = z ∧ y = w) ∧ φ))
9	1, 8	bitri 240	1 ⊢ ([z / x][w / y]φ ↔ ∃x∃y((x = z ∧ y = w) ∧ 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:  pm11.07  2115