Theorem 0funcglem 50160

Description: Lemma for 0funcg 50162. (Contributed by Zhi Wang, 17-Oct-2025.)

Hypotheses

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

Assertion

Ref	Expression
0funcglem	⊢ (𝜑 → (𝜓 ↔ (𝜂 ∧ 𝜁)))

Proof of Theorem 0funcglem

Step	Hyp	Ref	Expression
1		0funcglem.4	. . 3 ⊢ (𝜑 → 𝜏)
2		0funcglem.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃 ∧ 𝜏)))
3		df-3an 1105	. . . 4 ⊢ ((𝜒 ∧ 𝜃 ∧ 𝜏) ↔ ((𝜒 ∧ 𝜃) ∧ 𝜏))
4	2, 3	bitrdi 290	. . 3 ⊢ (𝜑 → (𝜓 ↔ ((𝜒 ∧ 𝜃) ∧ 𝜏)))
5	1, 4	mpbiran2d 721	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))
6		0funcglem.2	. . 3 ⊢ (𝜑 → (𝜒 ↔ 𝜂))
7		0funcglem.3	. . 3 ⊢ (𝜑 → (𝜃 ↔ 𝜁))
8	6, 7	anbi12d 644	. 2 ⊢ (𝜑 → ((𝜒 ∧ 𝜃) ↔ (𝜂 ∧ 𝜁)))
9	5, 8	bitrd 282	1 ⊢ (𝜑 → (𝜓 ↔ (𝜂 ∧ 𝜁)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105

This theorem is used by:  0funcg2  50161