Metamath Proof Explorer


Theorem 0funclem

Description: Lemma for 0funcALT . (Contributed by Zhi Wang, 7-Oct-2025)

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

Proof

Step Hyp Ref Expression
1 0funclem.1 ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ∧ 𝜏 ) ) )
2 0funclem.2 ⊢ ( 𝜒 ↔ 𝜂 )
3 0funclem.3 ⊢ ( 𝜃 ↔ 𝜁 )
4 0funclem.4 ⊢ 𝜏
5 df-3an ⊢ ( ( 𝜒 ∧ 𝜃 ∧ 𝜏 ) ↔ ( ( 𝜒 ∧ 𝜃 ) ∧ 𝜏 ) )
6 1 5 bitrdi ⊢ ( 𝜑 → ( 𝜓 ↔ ( ( 𝜒 ∧ 𝜃 ) ∧ 𝜏 ) ) )
7 6 rbaibd ⊢ ( ( 𝜑 ∧ 𝜏 ) → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) )
8 4 7 mpan2 ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜒 ∧ 𝜃 ) ) )
9 2 3 anbi12i ⊢ ( ( 𝜒 ∧ 𝜃 ) ↔ ( 𝜂 ∧ 𝜁 ) )
10 8 9 bitrdi ⊢ ( 𝜑 → ( 𝜓 ↔ ( 𝜂 ∧ 𝜁 ) ) )