Theorem friOLD 5599

Description: Obsolete version of fri 5598 as of 16-Nov-2024. (Contributed by NM, 18-Mar-1997.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
friOLD	⊢ (((𝐵 ∈ 𝐶 ∧ 𝑅 Fr 𝐴) ∧ (𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅)) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)

Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦   𝑥,𝑅,𝑦

Allowed substitution hints:   𝐶(𝑥,𝑦)

Proof of Theorem friOLD

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-fr 5593	. . 3 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑧((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥))
2		sseq1 3972	. . . . . 6 ⊢ (𝑧 = 𝐵 → (𝑧 ⊆ 𝐴 ↔ 𝐵 ⊆ 𝐴))
3		neeq1 3002	. . . . . 6 ⊢ (𝑧 = 𝐵 → (𝑧 ≠ ∅ ↔ 𝐵 ≠ ∅))
4	2, 3	anbi12d 631	. . . . 5 ⊢ (𝑧 = 𝐵 → ((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅)))
5		raleq 3307	. . . . . 6 ⊢ (𝑧 = 𝐵 → (∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥 ↔ ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥))
6	5	rexeqbi1dv 3306	. . . . 5 ⊢ (𝑧 = 𝐵 → (∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥 ↔ ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥))
7	4, 6	imbi12d 344	. . . 4 ⊢ (𝑧 = 𝐵 → (((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥) ↔ ((𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)))
8	7	spcgv 3556	. . 3 ⊢ (𝐵 ∈ 𝐶 → (∀𝑧((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥) → ((𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)))
9	1, 8	biimtrid 241	. 2 ⊢ (𝐵 ∈ 𝐶 → (𝑅 Fr 𝐴 → ((𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)))
10	9	imp31 418	1 ⊢ (((𝐵 ∈ 𝐶 ∧ 𝑅 Fr 𝐴) ∧ (𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅)) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396  ∀wal 1539   = wceq 1541   ∈ wcel 2106   ≠ wne 2939  ∀wral 3060  ∃wrex 3069   ⊆ wss 3913  ∅c0 4287   class class class wbr 5110   Fr wfr 5590

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 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2702

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2709  df-cleq 2723  df-clel 2809  df-ne 2940  df-ral 3061  df-rex 3070  df-v 3448  df-in 3920  df-ss 3930  df-fr 5593

This theorem is referenced by: (None)