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Theorem sb3an 2080
Description: Threefold 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 1083 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
21sbbii 2074 . 2 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ [𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒))
3 sban 2079 . 2 ([𝑦 / 𝑥]((𝜑𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒))
4 sban 2079 . . . 4 ([𝑦 / 𝑥](𝜑𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓))
54anbi1i 623 . . 3 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
6 df-3an 1083 . . 3 (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒))
75, 6bitr4i 279 . 2 (([𝑦 / 𝑥](𝜑𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
82, 3, 73bitri 298 1 ([𝑦 / 𝑥](𝜑𝜓𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 207  wa 396  w3a 1081  [wsb 2062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803
This theorem depends on definitions:  df-bi 208  df-an 397  df-3an 1083  df-sb 2063
This theorem is referenced by: (None)
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