Theorem sbanvOLD 2325

Description: Obsolete version of sban 2085 as of 24-Jul-2023. Substitution distributes over conjunction. Version of sban 2085 with a disjoint variable condition, not requiring ax-13 2389. (Contributed by Wolf Lammen, 18-Jan-2023.) (New usage is discouraged.) (Proof modification is discouraged.)

Assertion

Ref	Expression
sbanvOLD	⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem sbanvOLD

Step	Hyp	Ref	Expression
1		sbn 2286	. . 3 ⊢ ([𝑦 / 𝑥] ¬ (𝜑 → ¬ 𝜓) ↔ ¬ [𝑦 / 𝑥](𝜑 → ¬ 𝜓))
2		sbimvOLD 2324	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 → ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥] ¬ 𝜓))
3		sbn 2286	. . . . 5 ⊢ ([𝑦 / 𝑥] ¬ 𝜓 ↔ ¬ [𝑦 / 𝑥]𝜓)
4	3	imbi2i 338	. . . 4 ⊢ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥] ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → ¬ [𝑦 / 𝑥]𝜓))
5	2, 4	bitri 277	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 → ¬ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → ¬ [𝑦 / 𝑥]𝜓))
6	1, 5	xchbinx 336	. 2 ⊢ ([𝑦 / 𝑥] ¬ (𝜑 → ¬ 𝜓) ↔ ¬ ([𝑦 / 𝑥]𝜑 → ¬ [𝑦 / 𝑥]𝜓))
7		df-an 399	. . 3 ⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 → ¬ 𝜓))
8	7	sbbii 2080	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑥] ¬ (𝜑 → ¬ 𝜓))
9		df-an 399	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ↔ ¬ ([𝑦 / 𝑥]𝜑 → ¬ [𝑦 / 𝑥]𝜓))
10	6, 8, 9	3bitr4i 305	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398  [wsb 2068

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1969  ax-7 2014  ax-10 2144  ax-12 2176

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ex 1780  df-nf 1784  df-sb 2069

This theorem is referenced by:  sbbivOLD  2326