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Theorem sb3an 2537
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 1074 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
21sbbii 2053 . 2 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ [𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒))
3 sban 2536 . 2 ([𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒))
4 sban 2536 . . . 4 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
54anbi1i 733 . . 3 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6 df-3an 1074 . . 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 383  w3a 1072  [wsb 2046
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1871  ax-4 1886  ax-5 1988  ax-6 2054  ax-7 2090  ax-10 2168  ax-12 2196  ax-13 2391
This theorem depends on definitions:  df-bi 197  df-or 384  df-an 385  df-3an 1074  df-ex 1854  df-nf 1859  df-sb 2047
This theorem is referenced by: (None)
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