Theorem opnneilem 49983

Description: Lemma factoring out common proof steps of opnneil 49987 and opnneirv 49985. (Contributed by Zhi Wang, 31-Aug-2024.)

Hypothesis

Ref	Expression
opnneilem.1	⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
opnneilem	⊢ (𝜑 → (∃𝑥 ∈ 𝐽 (𝑆 ⊆ 𝑥 ∧ 𝜓) ↔ ∃𝑦 ∈ 𝐽 (𝑆 ⊆ 𝑦 ∧ 𝜒)))

Distinct variable groups:   𝑥,𝐽,𝑦   𝑥,𝑆,𝑦   𝜒,𝑥   𝜑,𝑥,𝑦   𝜓,𝑦

Allowed substitution hints:   𝜓(𝑥)   𝜒(𝑦)

Proof of Theorem opnneilem

Step	Hyp	Ref	Expression
1		sseq2 3957	. . . 4 ⊢ (𝑥 = 𝑦 → (𝑆 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑦))
2	1	adantl 487	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝑆 ⊆ 𝑥 ↔ 𝑆 ⊆ 𝑦))
3		opnneilem.1	. . 3 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → (𝜓 ↔ 𝜒))
4	2, 3	anbi12d 644	. 2 ⊢ ((𝜑 ∧ 𝑥 = 𝑦) → ((𝑆 ⊆ 𝑥 ∧ 𝜓) ↔ (𝑆 ⊆ 𝑦 ∧ 𝜒)))
5	4	cbvrexdva 3244	1 ⊢ (𝜑 → (∃𝑥 ∈ 𝐽 (𝑆 ⊆ 𝑥 ∧ 𝜓) ↔ ∃𝑦 ∈ 𝐽 (𝑆 ⊆ 𝑦 ∧ 𝜒)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∃wrex 3087   ⊆ wss 3899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-ss 3916

This theorem is used by:  opnneirv  49985  opnneil  49987  opnneibid2  49989