Theorem sb6x 2029

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypothesis

Ref	Expression
sb6x.1	⊢ Ⅎxφ

Assertion

Ref	Expression
sb6x	⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 ⊢ Ⅎxφ
2	1	sbf 2026	. 2 ⊢ ([y / x]φ ↔ φ)
3		biidd 228	. . 3 ⊢ (x = y → (φ ↔ φ))
4	1, 3	equsal 1960	. 2 ⊢ (∀x(x = y → φ) ↔ φ)
5	2, 4	bitr4i 243	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 4   ↔ wb 176  ∀wal 1540  Ⅎwnf 1544  [wsb 1648

This theorem was proved from 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 theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by: (None)