Theorem sb6rfv 2389

Description: Reversed substitution. Version of sb6rf 2500 requiring disjoint variables, but fewer axioms. (Contributed by NM, 1-Aug-1993.) (Revised by Wolf Lammen, 7-Feb-2023.)

Hypothesis

Ref	Expression
sb6rfv.nf	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
sb6rfv	⊢ (𝜑 ↔ ∀𝑦(𝑦 = 𝑥 → [𝑦 / 𝑥]𝜑))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem sb6rfv

Step	Hyp	Ref	Expression
1		sb6rfv.nf	. . 3 ⊢ Ⅎ𝑦𝜑
2		sbequ12r 2288	. . 3 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
3	1, 2	equsalv 2303	. 2 ⊢ (∀𝑦(𝑦 = 𝑥 → [𝑦 / 𝑥]𝜑) ↔ 𝜑)
4	3	bicomi 227	1 ⊢ (𝜑 ↔ ∀𝑦(𝑦 = 𝑥 → [𝑦 / 𝑥]𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  Ⅎwnf 1813  [wsb 2096

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-12 2213

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-nf 1814  df-sb 2097

This theorem is used by:  eu1  2638