Theorem sb5 2100

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40. (Contributed by NM, 18-Aug-1993.)

Assertion

Ref	Expression
sb5	⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

Distinct variable group:   x,y

Allowed substitution hints:   φ(x,y)

Proof of Theorem sb5

Step	Hyp	Ref	Expression
1		sb6 2099	. 2 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))
2		sb56 2098	. 2 ⊢ (∃x(x = y ∧ φ) ↔ ∀x(x = y → φ))
3	1, 2	bitr4i 243	1 ⊢ ([y / x]φ ↔ ∃x(x = y ∧ φ))

Syntax hints:   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541  [wsb 1648

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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:  2sb5  2112  dfsb7  2119  sb7f  2120  sbelx  2124  sbc2or  3054  sbc5  3070