Theorem sbelx 2267

Description: Elimination of substitution. Also see sbel2x 2484. (Contributed by NM, 5-Aug-1993.) Avoid ax-13 2382. (Revised by Wolf Lammen, 6-Aug-2023.) Avoid ax-10 2154. (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 2266	. . 3 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑦]𝜑 ↔ 𝜑))
2	1	equsexvw 2013	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑) ↔ 𝜑)
3	2	bicomi 226	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Syntax hints:   ↔ wb 208   ∧ wa 397  ∃wex 1787  [wsb 2074

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-12 2191

This theorem depends on definitions:  df-bi 209  df-an 398  df-ex 1788  df-sb 2075

This theorem is referenced by:  pm13.196a  44871