Theorem sb5 1899

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 1898	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
2		sb56 1897	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	bitr4i 187	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1362  ∃wex 1503  [wsb 1773

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 1458  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545

This theorem depends on definitions:  df-bi 117  df-sb 1774

This theorem is referenced by:  sbnv  1900  sborv  1902  sbi2v  1904  nfsbxy  1958  nfsbxyt  1959  2sb5  1999  dfsb7  2007  sb7f  2008  sbexyz  2019  sbc5  3009