Theorem sbor 2307

Description: Disjunction inside and outside of a substitution are equivalent. (Contributed by NM, 29-Sep-2002.)

Assertion

Ref	Expression
sbor	⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Proof of Theorem sbor

Step	Hyp	Ref	Expression
1		sbim 2303	. . 3 ⊢ ([𝑦 / 𝑥](¬ 𝜑 → 𝜓) ↔ ([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓))
2		sbn 2280	. . . 4 ⊢ ([𝑦 / 𝑥] ¬ 𝜑 ↔ ¬ [𝑦 / 𝑥]𝜑)
3	2	imbi1i 349	. . 3 ⊢ (([𝑦 / 𝑥] ¬ 𝜑 → [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
4	1, 3	bitri 274	. 2 ⊢ ([𝑦 / 𝑥](¬ 𝜑 → 𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
5		df-or 844	. . 3 ⊢ ((𝜑 ∨ 𝜓) ↔ (¬ 𝜑 → 𝜓))
6	5	sbbii 2080	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](¬ 𝜑 → 𝜓))
7		df-or 844	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ (¬ [𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
8	4, 6, 7	3bitr4i 302	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∨ wo 843  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-10 2139  ax-12 2173

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-ex 1784  df-nf 1788  df-sb 2069

This theorem is referenced by:  sbcor  3764  unab  4229