Theorem friOLD 5561

Description: Obsolete version of fri 5560 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 5555	. . 3 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑧((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥))
2		sseq1 3951	. . . . . 6 ⊢ (𝑧 = 𝐵 → (𝑧 ⊆ 𝐴 ↔ 𝐵 ⊆ 𝐴))
3		neeq1 3004	. . . . . 6 ⊢ (𝑧 = 𝐵 → (𝑧 ≠ ∅ ↔ 𝐵 ≠ ∅))
4	2, 3	anbi12d 632	. . . . 5 ⊢ (𝑧 = 𝐵 → ((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) ↔ (𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅)))
5		raleq 3361	. . . . . 6 ⊢ (𝑧 = 𝐵 → (∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥 ↔ ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥))
6	5	rexeqbi1dv 3360	. . . . 5 ⊢ (𝑧 = 𝐵 → (∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥 ↔ ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥))
7	4, 6	imbi12d 345	. . . 4 ⊢ (𝑧 = 𝐵 → (((𝑧 ⊆ 𝐴 ∧ 𝑧 ≠ ∅) → ∃𝑥 ∈ 𝑧 ∀𝑦 ∈ 𝑧 ¬ 𝑦𝑅𝑥) ↔ ((𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)))
8	7	spcgv 3540	. . 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 419	1 ⊢ (((𝐵 ∈ 𝐶 ∧ 𝑅 Fr 𝐴) ∧ (𝐵 ⊆ 𝐴 ∧ 𝐵 ≠ ∅)) → ∃𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ¬ 𝑦𝑅𝑥)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 397  ∀wal 1537   = wceq 1539   ∈ wcel 2104   ≠ wne 2941  ∀wral 3062  ∃wrex 3071   ⊆ wss 3892  ∅c0 4262   class class class wbr 5081   Fr wfr 5552

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2707

This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2714  df-cleq 2728  df-clel 2814  df-ne 2942  df-ral 3063  df-rex 3072  df-v 3439  df-in 3899  df-ss 3909  df-fr 5555

This theorem is referenced by: (None)