Theorem sb5 2300

Description: Equivalence for substitution. Similar to Theorem 6.1 of [Quine] p. 40. The implication "to the right" is sb1 2067 and does not require any disjoint variable condition. Theorem sb5f 2503 replaces the disjoint variable condition with a non-freeness hypothesis. (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 2299	. 2 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
2		sb56 2297	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑))
3	1, 2	bitr4i 270	1 ⊢ ([𝑦 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 385  ∀wal 1651  ∃wex 1875  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-10 2185  ax-12 2213

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-ex 1876  df-nf 1880  df-sb 2065

This theorem is referenced by:  2sb5  2301  sbnv  2326  sb7f  2573  sbelx  2577  sbc2or  3642  sbc5  3658