Theorem oppfrcllem 49933

Description: Lemma for oppfrcl 49934. (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 5720	. . 3 ⊢ (Rel 𝑅 → ¬ ∅ ∈ 𝑅)
4	2, 3	ax-mp 5	. 2 ⊢ ¬ ∅ ∈ 𝑅
5		nelne2 3055	. 2 ⊢ ((𝐺 ∈ 𝑅 ∧ ¬ ∅ ∈ 𝑅) → 𝐺 ≠ ∅)
6	1, 4, 5	sylancl 597	1 ⊢ (𝜑 → 𝐺 ≠ ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∈ wcel 2142   ≠ wne 2957  ∅c0 4285  Rel wrel 5665

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-sep 5256  ax-pr 5403

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-opab 5173  df-xp 5666  df-rel 5667

This theorem is used by:  oppfrcl  49934  oppfrcl3  49936  lmdran  50477  cmdlan  50478