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Theorem sb3an 2399
 Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)
Assertion
Ref Expression
sb3an ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))

Proof of Theorem sb3an
StepHypRef Expression
1 df-3an 1039 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
21sbbii 1886 . 2 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ [𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒))
3 sban 2398 . 2 ([𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒))
4 sban 2398 . . . 4 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
54anbi1i 731 . . 3 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6 df-3an 1039 . . 3 (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
75, 6bitr4i 267 . 2 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
82, 3, 73bitri 286 1 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 196   ∧ wa 384   ∧ w3a 1037  [wsb 1879 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1721  ax-4 1736  ax-5 1838  ax-6 1887  ax-7 1934  ax-10 2018  ax-12 2046  ax-13 2245 This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-ex 1704  df-nf 1709  df-sb 1880 This theorem is referenced by: (None)
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