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Theorem sbor 2308
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 2304 . . 3 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ ([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓))
2 sbn 2281 . . . 4 ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
32imbi1i 350 . . 3 (([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
41, 3bitri 274 . 2 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
5 df-or 845 . . 3 ((𝜑𝜓) ↔ (¬ 𝜑𝜓))
65sbbii 2083 . 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 303 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wo 844  [wsb 2071
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-10 2141  ax-12 2175
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1787  df-nf 1791  df-sb 2072
This theorem is referenced by:  sbcor  3773  unab  4238
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