Theorem sb5 1880

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40. (Contributed by NM, 18-Aug-1993.) (Revised by NM, 14-Apr-2008.)

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 1879	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
2		sb56 1878	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	bitr4i 186	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1346  ∃wex 1485  [wsb 1755

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 1440  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527

This theorem depends on definitions:  df-bi 116  df-sb 1756

This theorem is referenced by:  sbnv  1881  sborv  1883  sbi2v  1885  nfsbxy  1935  nfsbxyt  1936  2sb5  1976  dfsb7  1984  sb7f  1985  sbexyz  1996  sbc5  2978