Theorem sepnsepolem1 49728

Description: Lemma for sepnsepo 49730. (Contributed by Zhi Wang, 1-Sep-2024.)

Assertion

Ref	Expression
sepnsepolem1	⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∃𝑥 ∈ 𝐽 (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))

Distinct variable group:   𝜑,𝑦

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥, 𝑦)   𝜒(𝑥, 𝑦)   𝐽(𝑥, 𝑦)

Proof of Theorem sepnsepolem1

Step	Hyp	Ref	Expression
1		3anass 1110	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (𝜑 ∧ (𝜓 ∧ 𝜒)))
2	1	2rexbii 3140	. 2 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)))
3		r19.42v 3196	. . 3 ⊢ (∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)) ↔ (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))
4	3	rexbii 3111	. 2 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)) ↔ ∃𝑥 ∈ 𝐽 (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))
5	2, 4	bitri 278	1 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∃𝑥 ∈ 𝐽 (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 400   ∧ w3a 1102  ∃wrex 3088

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939

This proof depends on definitions:  df-bi 210  df-an 401  df-3an 1104  df-ex 1809  df-rex 3089

This theorem is used by:  sepnsepo  49730