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Mirrors > Home > ILE Home > Th. List > 3jao | GIF version |
Description: Disjunction of 3 antecedents. (Contributed by NM, 8-Apr-1994.) |
Ref | Expression |
---|---|
3jao | ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-3or 886 | . 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜃) ↔ ((𝜑 ∨ 𝜒) ∨ 𝜃)) | |
2 | jao 672 | . . . 4 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜑 ∨ 𝜒) → 𝜓))) | |
3 | jao 672 | . . . 4 ⊢ (((𝜑 ∨ 𝜒) → 𝜓) → ((𝜃 → 𝜓) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓))) | |
4 | 2, 3 | syl6 29 | . . 3 ⊢ ((𝜑 → 𝜓) → ((𝜒 → 𝜓) → ((𝜃 → 𝜓) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓)))) |
5 | 4 | 3imp 1098 | . 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → (((𝜑 ∨ 𝜒) ∨ 𝜃) → 𝜓)) |
6 | 1, 5 | syl5bi 141 | 1 ⊢ (((𝜑 → 𝜓) ∧ (𝜒 → 𝜓) ∧ (𝜃 → 𝜓)) → ((𝜑 ∨ 𝜒 ∨ 𝜃) → 𝜓)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∨ wo 629 ∨ w3o 884 ∧ w3a 885 |
This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-io 630 |
This theorem depends on definitions: df-bi 110 df-3or 886 df-3an 887 |
This theorem is referenced by: 3jaob 1197 3jaoi 1198 3jaod 1199 |
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