Theorem sbelx 2026

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		ax-17 1550	. 2 ⊢ (𝜑 → ∀𝑥𝜑)
2	1	sb5rf 1876	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Syntax hints:   ∧ wa 104   ↔ wb 105  ∃wex 1516  [wsb 1786

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558

This theorem depends on definitions:  df-bi 117  df-sb 1787

This theorem is referenced by:  sbel2x  2027