Theorem 2sb6 2113

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

Assertion

Ref	Expression
2sb6	⊢ ([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 2sb6

Step	Hyp	Ref	Expression
1		sb6 2099	. 2 ⊢ ([z / x][w / y]φ ↔ ∀x(x = z → [w / y]φ))
2		19.21v 1890	. . . 4 ⊢ (∀y(x = z → (y = w → φ)) ↔ (x = z → ∀y(y = w → φ)))
3		impexp 433	. . . . 5 ⊢ (((x = z ∧ y = w) → φ) ↔ (x = z → (y = w → φ)))
4	3	albii 1566	. . . 4 ⊢ (∀y((x = z ∧ y = w) → φ) ↔ ∀y(x = z → (y = w → φ)))
5		sb6 2099	. . . . 5 ⊢ ([w / y]φ ↔ ∀y(y = w → φ))
6	5	imbi2i 303	. . . 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	albii 1566	. 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:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  [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:  2eu6  2289