Theorem sb6rf 2500

Description: Reversed substitution. For a version requiring disjoint variables, but fewer axioms, see sb6rfv 2389. Usage of this theorem is discouraged because it depends on ax-13 2404. Use the weaker sb6rfv 2389 if possible. (Contributed by NM, 1-Aug-1993.) (Revised by Mario Carneiro, 6-Oct-2016.) (Proof shortened by Wolf Lammen, 21-Sep-2018.) (New usage is discouraged.)

Hypothesis

Ref	Expression
sb5rf.1	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
sb6rf	⊢ (𝜑 ↔ ∀𝑦(𝑦 = 𝑥 → [𝑦 / 𝑥]𝜑))

Proof of Theorem sb6rf

Step	Hyp	Ref	Expression
1		sb5rf.1	. . 3 ⊢ Ⅎ𝑦𝜑
2		sbequ12r 2288	. . 3 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
3	1, 2	equsal 2449	. 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  ax-13 2404

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: (None)