Theorem oppfrcllem 49825

Description: Lemma for oppfrcl 49826. (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 5722	. . 3 ⊢ (Rel 𝑅 → ¬ ∅ ∈ 𝑅)
4	2, 3	ax-mp 5	. 2 ⊢ ¬ ∅ ∈ 𝑅
5		nelne2 3062	. 2 ⊢ ((𝐺 ∈ 𝑅 ∧ ¬ ∅ ∈ 𝑅) → 𝐺 ≠ ∅)
6	1, 4, 5	sylancl 597	1 ⊢ (𝜑 → 𝐺 ≠ ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∈ wcel 2149   ≠ wne 2964  ∅c0 4292  Rel wrel 5667

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741  ax-sep 5259  ax-pr 5405

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ne 2965  df-rab 3423  df-v 3463  df-dif 3914  df-un 3916  df-in 3918  df-ss 3928  df-nul 4293  df-if 4491  df-sn 4593  df-pr 4595  df-op 4599  df-opab 5176  df-xp 5668  df-rel 5669

This theorem is referenced by:  oppfrcl  49826  oppfrcl3  49828  lmdran  50369  cmdlan  50370