Theorem oppfrcllem 50157

Description: Lemma for oppfrcl 50158. (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 5707	. . 3 ⊢ (Rel 𝑅 → ¬ ∅ ∈ 𝑅)
4	2, 3	ax-mp 5	. 2 ⊢ ¬ ∅ ∈ 𝑅
5		nelne2 3053	. 2 ⊢ ((𝐺 ∈ 𝑅 ∧ ¬ ∅ ∈ 𝑅) → 𝐺 ≠ ∅)
6	1, 4, 5	sylancl 598	1 ⊢ (𝜑 → 𝐺 ≠ ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∈ wcel 2145   ≠ wne 2955  ∅c0 4278  Rel wrel 5652

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 2732  ax-sep 5248  ax-pr 5390

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ne 2956  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-opab 5167  df-xp 5653  df-rel 5654

This theorem is used by:  oppfrcl  50158  oppfrcl3  50160  lmdran  50701  cmdlan  50702