Theorem oppfrcllem 49122

Description: Lemma for oppfrcl 49123. (Contributed by Zhi Wang, 14-Nov-2025.)

Hypotheses

Ref	Expression
oppfrcl.1	⊢ (𝜑 → 𝐺 ∈ 𝑅)
oppfrcl.2	⊢ Rel 𝑅

Assertion

Ref	Expression
oppfrcllem	⊢ (𝜑 → 𝐺 ≠ ∅)

Proof of Theorem oppfrcllem

Step	Hyp	Ref	Expression
1		oppfrcl.1	. 2 ⊢ (𝜑 → 𝐺 ∈ 𝑅)
2		oppfrcl.2	. . 3 ⊢ Rel 𝑅
3		0nelrel0 5679	. . 3 ⊢ (Rel 𝑅 → ¬ ∅ ∈ 𝑅)
4	2, 3	ax-mp 5	. 2 ⊢ ¬ ∅ ∈ 𝑅
5		nelne2 3023	. 2 ⊢ ((𝐺 ∈ 𝑅 ∧ ¬ ∅ ∈ 𝑅) → 𝐺 ≠ ∅)
6	1, 4, 5	sylancl 586	1 ⊢ (𝜑 → 𝐺 ≠ ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∈ wcel 2109   ≠ wne 2925  ∅c0 4284  Rel wrel 5624

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5235  ax-nul 5245  ax-pr 5371

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-ne 2926  df-v 3438  df-dif 3906  df-un 3908  df-ss 3920  df-nul 4285  df-if 4477  df-sn 4578  df-pr 4580  df-op 4584  df-opab 5155  df-xp 5625  df-rel 5626

This theorem is referenced by:  oppfrcl  49123  oppfrcl3  49125  lmdran  49666  cmdlan  49667