Theorem sb3an 2086

Description: Threefold 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 1086	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2	1	sbbii 2081	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒))
3		sban 2085	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒))
4		sban 2085	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
5	4	anbi1i 626	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6		df-3an 1086	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
7	5, 6	bitr4i 281	. 2 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
8	2, 3, 7	3bitri 300	1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))

Syntax hints:   ↔ wb 209   ∧ wa 399   ∧ w3a 1084  [wsb 2069

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 210  df-an 400  df-3an 1086  df-sb 2070

This theorem is referenced by: (None)