Metamath Proof Explorer


Theorem 0funcglem

Description: Lemma for 0funcg . (Contributed by Zhi Wang, 17-Oct-2025)

Ref Expression
Hypotheses 0funcglem.1 ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ∧ 𝜏 ) ) )
0funcglem.2 ⊢ ( 𝜑 → ( 𝜒 ↔ 𝜂 ) )
0funcglem.3 ⊢ ( 𝜑 → ( 𝜃 ↔ 𝜁 ) )
0funcglem.4 ⊢ ( 𝜑 → 𝜏 )
Assertion 0funcglem ( 𝜑 → ( 𝜓 ↔ ( 𝜂 ∧ 𝜁 ) ) )

Proof

Step Hyp Ref Expression
1 0funcglem.1 ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ∧ 𝜏 ) ) )
2 0funcglem.2 ⊢ ( 𝜑 → ( 𝜒 ↔ 𝜂 ) )
3 0funcglem.3 ⊢ ( 𝜑 → ( 𝜃 ↔ 𝜁 ) )
4 0funcglem.4 ⊢ ( 𝜑 → 𝜏 )
5 df-3an ⊢ ( ( 𝜒 ∧ 𝜃 ∧ 𝜏 ) ↔ ( ( 𝜒 ∧ 𝜃 ) ∧ 𝜏 ) )
6 1 5 bitrdi ⊢ ( 𝜑 → ( 𝜓 ↔ ( ( 𝜒 ∧ 𝜃 ) ∧ 𝜏 ) ) )
7 4 6 mpbiran2d ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) )
8 2 3 anbi12d ⊢ ( 𝜑 → ( ( 𝜒 ∧ 𝜃 ) ↔ ( 𝜂 ∧ 𝜁 ) ) )
9 7 8 bitrd ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜂 ∧ 𝜁 ) ) )