Theorem sbelx 2288

Description: Elimination of substitution. Also see sbel2x 2505. (Contributed by NM, 5-Aug-1993.) Avoid ax-13 2403. (Revised by Wolf Lammen, 6-Aug-2023.) Avoid ax-10 2175. (Revised by GG, 20-Aug-2023.)

Assertion

Ref	Expression
sbelx	⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Distinct variable groups:   𝑥,𝑦   𝜑,𝑥

Allowed substitution hint:   𝜑(𝑦)

Proof of Theorem sbelx

Step	Hyp	Ref	Expression
1		sbequ12r 2287	. . 3 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑 ↔ 𝜑))
2	1	equsexvw 2034	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑) ↔ 𝜑)
3	2	bicomi 227	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 400  ∃wex 1808  [wsb 2095

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-sb 2096

This theorem is used by:  pm13.196a  45152