Theorem sbelx 2181

Description: Elimination of substitution. Also see sbel2x 2421. (Contributed by NM, 5-Aug-1993.) Avoid ax-13 2301. (Revised by Wolf Lammen, 6-Aug-2023.) Avoid ax-10 2079. (Revised by Gino Giotto, 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 2180	. . 3 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑 ↔ 𝜑))
2	1	equsexvw 1962	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑) ↔ 𝜑)
3	2	bicomi 216	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Syntax hints:   ↔ wb 198   ∧ wa 387  ∃wex 1742  [wsb 2015

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-12 2106

This theorem depends on definitions:  df-bi 199  df-an 388  df-ex 1743  df-sb 2016

This theorem is referenced by:  pm13.196a  40163