Theorem sbelx 2486

Description: Elimination of substitution. (Contributed by NM, 5-Aug-1993.)

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		sbid2v 2485	. 2 ⊢ ([𝑦 / 𝑥][𝑥 / 𝑦]𝜑 ↔ 𝜑)
2		sb5 2458	. 2 ⊢ ([𝑦 / 𝑥][𝑥 / 𝑦]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))
3	1, 2	bitr3i 266	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Syntax hints:   ↔ wb 196   ∧ wa 383  ∃wex 1744  [wsb 1937

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1762  ax-4 1777  ax-5 1879  ax-6 1945  ax-7 1981  ax-10 2059  ax-12 2087  ax-13 2282

This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-ex 1745  df-nf 1750  df-sb 1938

This theorem is referenced by:  pm13.196a  38932