Theorem sb6x 1767

Description: Equivalence involving substitution for a variable not free. (Contributed by NM, 5-Aug-1993.) (Proof shortened by Andrew Salmon, 12-Aug-2011.)

Hypothesis

Ref	Expression
sb6x.1	⊢ (𝜑 → ∀𝑥𝜑)

Assertion

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

Proof of Theorem sb6x

Step	Hyp	Ref	Expression
1		sb6x.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2	1	sbh 1764	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		biidd 171	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜑))
4	1, 3	equsalh 1714	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
5	2, 4	bitr4i 186	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1341  [wsb 1750

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-4 1498  ax-i9 1518  ax-ial 1522

This theorem depends on definitions:  df-bi 116  df-sb 1751

This theorem is referenced by: (None)