Theorem sb56 2098

Description: Two equivalent ways of expressing the proper substitution of y for x in φ, when x and y are distinct. Theorem 6.2 of [Quine] p. 40. The proof does not involve df-sb 1649. (Contributed by NM, 14-Apr-2008.)

Assertion

Ref	Expression
sb56	⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))

Distinct variable group:   x,y

Allowed substitution hints:   φ(x, y)

Proof of Theorem sb56

Step	Hyp	Ref	Expression
1		nfa1 1788	. 2 ⊢ Ⅎx∀x(x = y → φ)
2		ax11v 2096	. . 3 ⊢ (x = y → (φ → ∀x(x = y → φ)))
3		sp 1747	. . . 4 ⊢ (∀x(x = y → φ) → (x = y → φ))
4	3	com12 27	. . 3 ⊢ (x = y → (∀x(x = y → φ) → φ))
5	2, 4	impbid 183	. 2 ⊢ (x = y → (φ ↔ ∀x(x = y → φ)))
6	1, 5	equsex 1962	1 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541

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

This theorem is used by:  sb6  2099  sb5  2100  alexeq  2969