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Theorem sbor 2312
 Description: Disjunction inside and outside of a substitution are equivalent. (Contributed by NM, 29-Sep-2002.)
Assertion
Ref Expression
sbor ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Proof of Theorem sbor
StepHypRef Expression
1 sbim 2307 . . 3 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ ([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓))
2 sbn 2283 . . . 4 ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
32imbi1i 353 . . 3 (([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
41, 3bitri 278 . 2 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
5 df-or 845 . . 3 ((𝜑𝜓) ↔ (¬ 𝜑𝜓))
65sbbii 2081 . 2 ([𝑦 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑥](¬ 𝜑𝜓))
7 df-or 845 . 2 (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
84, 6, 73bitr4i 306 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∨ wo 844  [wsb 2069 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-10 2142  ax-12 2175 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-nf 1786  df-sb 2070 This theorem is referenced by:  sbcor  3769  unab  4222
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