Theorem sbelx 1991

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 1520	. 2 ⊢ (𝜑 → ∀𝑥𝜑)
2	1	sb5rf 1846	1 ⊢ (𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ [𝑥 / 𝑦]𝜑))

Syntax hints:   ∧ wa 103   ↔ wb 104  ∃wex 1486  [wsb 1756

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1441  ax-gen 1443  ax-ie1 1487  ax-ie2 1488  ax-8 1498  ax-11 1500  ax-4 1504  ax-17 1520  ax-i9 1524  ax-ial 1528

This theorem depends on definitions:  df-bi 116  df-sb 1757

This theorem is referenced by:  sbel2x  1992