Theorem sb6 2099

Description: Equivalence for substitution. Compare Theorem 6.2 of [Quine] p. 40. Also proved as Lemmas 16 and 17 of [Tarski] p. 70. (Contributed by NM, 18-Aug-1993.)

Assertion

Ref	Expression
sb6	⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Distinct variable group:   x,y

Allowed substitution hints:   φ(x, y)

Proof of Theorem sb6

Step	Hyp	Ref	Expression
1		sb56 2098	. . 3 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))
2	1	anbi2i 675	. 2 ⊢ (((x = y → φ) ∧ ∃x(x = y ∧ φ)) ↔ ((x = y → φ) ∧ ∀x(x = y → φ)))
3		df-sb 1649	. 2 ⊢ ([y / x]φ ↔ ((x = y → φ) ∧ ∃x(x = y ∧ φ)))
4		sp 1747	. . 3 ⊢ (∀x(x = y → φ) → (x = y → φ))
5	4	pm4.71ri 614	. 2 ⊢ (∀x(x = y → φ) ↔ ((x = y → φ) ∧ ∀x(x = y → φ)))
6	2, 3, 5	3bitr4i 268	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃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:  sb5  2100  2sb6  2113  sb6a  2116  exsbOLD  2131  sbal2  2134