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Mirrors > Home > ILE Home > Th. List > Mathboxes > bj-stand | GIF version |
Description: The conjunction of two stable formulas is stable. Deduction form of bj-stan 13628. Its proof is shorter (when counting all steps, including syntactic steps), so one could prove it first and then bj-stan 13628 from it, the usual way. (Contributed by BJ, 24-Nov-2023.) (Proof modification is discouraged.) |
Ref | Expression |
---|---|
bj-stand.1 | ⊢ (𝜑 → STAB 𝜓) |
bj-stand.2 | ⊢ (𝜑 → STAB 𝜒) |
Ref | Expression |
---|---|
bj-stand | ⊢ (𝜑 → STAB (𝜓 ∧ 𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bj-nnan 13617 | . . 3 ⊢ (¬ ¬ (𝜓 ∧ 𝜒) → (¬ ¬ 𝜓 ∧ ¬ ¬ 𝜒)) | |
2 | bj-stand.1 | . . . . 5 ⊢ (𝜑 → STAB 𝜓) | |
3 | df-stab 821 | . . . . 5 ⊢ (STAB 𝜓 ↔ (¬ ¬ 𝜓 → 𝜓)) | |
4 | 2, 3 | sylib 121 | . . . 4 ⊢ (𝜑 → (¬ ¬ 𝜓 → 𝜓)) |
5 | bj-stand.2 | . . . . 5 ⊢ (𝜑 → STAB 𝜒) | |
6 | df-stab 821 | . . . . 5 ⊢ (STAB 𝜒 ↔ (¬ ¬ 𝜒 → 𝜒)) | |
7 | 5, 6 | sylib 121 | . . . 4 ⊢ (𝜑 → (¬ ¬ 𝜒 → 𝜒)) |
8 | 4, 7 | anim12d 333 | . . 3 ⊢ (𝜑 → ((¬ ¬ 𝜓 ∧ ¬ ¬ 𝜒) → (𝜓 ∧ 𝜒))) |
9 | 1, 8 | syl5 32 | . 2 ⊢ (𝜑 → (¬ ¬ (𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒))) |
10 | df-stab 821 | . 2 ⊢ (STAB (𝜓 ∧ 𝜒) ↔ (¬ ¬ (𝜓 ∧ 𝜒) → (𝜓 ∧ 𝜒))) | |
11 | 9, 10 | sylibr 133 | 1 ⊢ (𝜑 → STAB (𝜓 ∧ 𝜒)) |
Colors of variables: wff set class |
Syntax hints: ¬ wn 3 → wi 4 ∧ wa 103 STAB wstab 820 |
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 604 ax-in2 605 |
This theorem depends on definitions: df-bi 116 df-stab 821 |
This theorem is referenced by: (None) |
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