Theorem 0funclem 50163

Description: Lemma for 0funcALT 50165. (Contributed by Zhi Wang, 7-Oct-2025.)

Hypotheses

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

Assertion

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

Proof of Theorem 0funclem

Step	Hyp	Ref	Expression
1		0funclem.4	. . 3 ⊢ 𝜏
2		0funclem.1	. . . . 5 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃 ∧ 𝜏)))
3		df-3an 1105	. . . . 5 ⊢ ((𝜒 ∧ 𝜃 ∧ 𝜏) ↔ ((𝜒 ∧ 𝜃) ∧ 𝜏))
4	2, 3	bitrdi 290	. . . 4 ⊢ (𝜑 → (𝜓 ↔ ((𝜒 ∧ 𝜃) ∧ 𝜏)))
5	4	rbaibd 550	. . 3 ⊢ ((𝜑 ∧ 𝜏) → (𝜓 ↔ (𝜒 ∧ 𝜃)))
6	1, 5	mpan2 704	. 2 ⊢ (𝜑 → (𝜓 ↔ (𝜒 ∧ 𝜃)))
7		0funclem.2	. . 3 ⊢ (𝜒 ↔ 𝜂)
8		0funclem.3	. . 3 ⊢ (𝜃 ↔ 𝜁)
9	7, 8	anbi12i 640	. 2 ⊢ ((𝜒 ∧ 𝜃) ↔ (𝜂 ∧ 𝜁))
10	6, 9	bitrdi 290	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:  0funcALT  50165