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Theorem bj-stand 13009
 Description: The conjunction of two stable formulas is stable. Deduction form of bj-stan 13008. Its proof is shorter, so one could prove it first and then bj-stan 13008 from it, the usual way. (Contributed by BJ, 24-Nov-2023.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-stand.1 (𝜑STAB 𝜓)
bj-stand.2 (𝜑STAB 𝜒)
Assertion
Ref Expression
bj-stand (𝜑STAB (𝜓𝜒))

Proof of Theorem bj-stand
StepHypRef Expression
1 bj-nnan 13001 . . 3 (¬ ¬ (𝜓𝜒) → (¬ ¬ 𝜓 ∧ ¬ ¬ 𝜒))
2 bj-stand.1 . . . . 5 (𝜑STAB 𝜓)
3 df-stab 816 . . . . 5 (STAB 𝜓 ↔ (¬ ¬ 𝜓𝜓))
42, 3sylib 121 . . . 4 (𝜑 → (¬ ¬ 𝜓𝜓))
5 bj-stand.2 . . . . 5 (𝜑STAB 𝜒)
6 df-stab 816 . . . . 5 (STAB 𝜒 ↔ (¬ ¬ 𝜒𝜒))
75, 6sylib 121 . . . 4 (𝜑 → (¬ ¬ 𝜒𝜒))
84, 7anim12d 333 . . 3 (𝜑 → ((¬ ¬ 𝜓 ∧ ¬ ¬ 𝜒) → (𝜓𝜒)))
91, 8syl5 32 . 2 (𝜑 → (¬ ¬ (𝜓𝜒) → (𝜓𝜒)))
10 df-stab 816 . 2 (STAB (𝜓𝜒) ↔ (¬ ¬ (𝜓𝜒) → (𝜓𝜒)))
119, 10sylibr 133 1 (𝜑STAB (𝜓𝜒))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  STAB wstab 815 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604 This theorem depends on definitions:  df-bi 116  df-stab 816 This theorem is referenced by: (None)
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