Theorem sb3an 1832

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)

Assertion

Ref	Expression
sb3an	⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))

Proof of Theorem sb3an

Step	Hyp	Ref	Expression
1		df-3an 887	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbbii 1648	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sban 1829	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒))
4		sban 1829	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
5	4	anbi1i 431	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6		df-3an 887	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
7	5, 6	bitr4i 176	. 2 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
8	2, 3, 7	3bitri 195	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))

Syntax hints:   ∧ wa 97   ↔ wb 98   ∧ w3a 885  [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428

This theorem depends on definitions:  df-bi 110  df-3an 887  df-nf 1350  df-sb 1646

This theorem is referenced by:  sbc3ang  2820