Theorem sb6x 2493

Description: Equivalence involving substitution for a variable not free. Usage of this theorem is discouraged because it depends on ax-13 2401. Usage of sb6 2121 is preferred, which requires fewer axioms. (Contributed by NM, 2-Jun-1993.) (Revised by Mario Carneiro, 4-Oct-2016.) (New usage is discouraged.)

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 2304	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜑)
3		biidd 265	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜑))
4	1, 3	equsal 2446	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜑)
5	2, 4	bitr4i 281	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  Ⅎwnf 1815  [wsb 2098

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-12 2212  ax-13 2401

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-sb 2099

This theorem is used by: (None)