Theorem oppfrcllem 49618

Description: Lemma for oppfrcl 49619. (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 5686	. . 3 ⊢ (Rel 𝑅 → ¬ ∅ ∈ 𝑅)
4	2, 3	ax-mp 5	. 2 ⊢ ¬ ∅ ∈ 𝑅
5		nelne2 3031	. 2 ⊢ ((𝐺 ∈ 𝑅 ∧ ¬ ∅ ∈ 𝑅) → 𝐺 ≠ ∅)
6	1, 4, 5	sylancl 587	1 ⊢ (𝜑 → 𝐺 ≠ ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∈ wcel 2114   ≠ wne 2933  ∅c0 4274  Rel wrel 5631

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709  ax-sep 5232  ax-pr 5372

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-opab 5149  df-xp 5632  df-rel 5633

This theorem is referenced by:  oppfrcl  49619  oppfrcl3  49621  lmdran  50162  cmdlan  50163