Theorem sepnsepolem1 46103

Description: Lemma for sepnsepo 46105. (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 1093	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (𝜑 ∧ (𝜓 ∧ 𝜒)))
2	1	2rexbii 3178	. 2 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)))
3		r19.42v 3276	. . 3 ⊢ (∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)) ↔ (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))
4	3	rexbii 3177	. 2 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ (𝜓 ∧ 𝜒)) ↔ ∃𝑥 ∈ 𝐽 (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))
5	2, 4	bitri 274	1 ⊢ (∃𝑥 ∈ 𝐽 ∃𝑦 ∈ 𝐽 (𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∃𝑥 ∈ 𝐽 (𝜑 ∧ ∃𝑦 ∈ 𝐽 (𝜓 ∧ 𝜒)))

Syntax hints:   ↔ wb 205   ∧ wa 395   ∧ w3a 1085  ∃wrex 3064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914

This theorem depends on definitions:  df-bi 206  df-an 396  df-3an 1087  df-ex 1784  df-rex 3069

This theorem is referenced by:  sepnsepo  46105