Description: Deduction joining 3 equivalences to form equivalence of disjunctions. (Contributed by NM, 20-Apr-1994)
Ref | Expression | ||
---|---|---|---|
Hypotheses | bi3d.1 | ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) | |
bi3d.2 | ⊢ ( 𝜑 → ( 𝜃 ↔ 𝜏 ) ) | ||
bi3d.3 | ⊢ ( 𝜑 → ( 𝜂 ↔ 𝜁 ) ) | ||
Assertion | 3orbi123d | ⊢ ( 𝜑 → ( ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) ↔ ( 𝜒 ∨ 𝜏 ∨ 𝜁 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | bi3d.1 | ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) | |
2 | bi3d.2 | ⊢ ( 𝜑 → ( 𝜃 ↔ 𝜏 ) ) | |
3 | bi3d.3 | ⊢ ( 𝜑 → ( 𝜂 ↔ 𝜁 ) ) | |
4 | 1 2 | orbi12d | ⊢ ( 𝜑 → ( ( 𝜓 ∨ 𝜃 ) ↔ ( 𝜒 ∨ 𝜏 ) ) ) |
5 | 4 3 | orbi12d | ⊢ ( 𝜑 → ( ( ( 𝜓 ∨ 𝜃 ) ∨ 𝜂 ) ↔ ( ( 𝜒 ∨ 𝜏 ) ∨ 𝜁 ) ) ) |
6 | df-3or | ⊢ ( ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) ↔ ( ( 𝜓 ∨ 𝜃 ) ∨ 𝜂 ) ) | |
7 | df-3or | ⊢ ( ( 𝜒 ∨ 𝜏 ∨ 𝜁 ) ↔ ( ( 𝜒 ∨ 𝜏 ) ∨ 𝜁 ) ) | |
8 | 5 6 7 | 3bitr4g | ⊢ ( 𝜑 → ( ( 𝜓 ∨ 𝜃 ∨ 𝜂 ) ↔ ( 𝜒 ∨ 𝜏 ∨ 𝜁 ) ) ) |