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Theorem sb3an 2387
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 1032 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
21sbbii 1873 . 2 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ [𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒))
3 sban 2386 . 2 ([𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒))
4 sban 2386 . . . 4 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
54anbi1i 726 . . 3 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6 df-3an 1032 . . 3 (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
75, 6bitr4i 265 . 2 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
82, 3, 73bitri 284 1 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 194  wa 382  w3a 1030  [wsb 1866
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-12 2033  ax-13 2233
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-3an 1032  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867
This theorem is referenced by: (None)
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