Theorem sbelx 2124

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 2123	. 2 ⊢ ([y / x][x / y]φ ↔ φ)
2		sb5 2100	. 2 ⊢ ([y / x][x / y]φ ↔ ∃x(x = y ∧ [x / y]φ))
3	1, 2	bitr3i 242	1 ⊢ (φ ↔ ∃x(x = y ∧ [x / y]φ))

Syntax hints:   ↔ wb 176   ∧ wa 358  ∃wex 1541  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sbel2x  2125