Theorem sb3an 1977

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 982	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbbii 1779	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sban 1974	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒))
4		sban 1974	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
5	4	anbi1i 458	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6		df-3an 982	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
7	5, 6	bitr4i 187	. 2 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
8	2, 3, 7	3bitri 206	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))

Syntax hints:   ∧ wa 104   ↔ wb 105   ∧ w3a 980  [wsb 1776

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549

This theorem depends on definitions:  df-bi 117  df-3an 982  df-nf 1475  df-sb 1777

This theorem is referenced by: (None)