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Theorem sbor 2556
Description: Logical OR inside and outside of 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 2553 . . 3 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ ([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓))
2 sbn 2549 . . . 4 ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
32imbi1i 340 . . 3 (([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
41, 3bitri 266 . 2 ([𝑦 / 𝑥](¬ 𝜑𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
5 df-or 866 . . 3 ((𝜑𝜓) ↔ (¬ 𝜑𝜓))
65sbbii 2066 . 2 ([𝑦 / 𝑥](𝜑𝜓) ↔ [𝑦 / 𝑥](¬ 𝜑𝜓))
7 df-or 866 . 2 (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
84, 6, 73bitr4i 294 1 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 197  wo 865  [wsb 2059
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2067  ax-7 2103  ax-10 2184  ax-12 2213  ax-13 2419
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-ex 1860  df-nf 1864  df-sb 2060
This theorem is referenced by:  sbcor  3671  unab  4089  sbcorgOLD  39232
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