Theorem prjsprel 43458

Description: Utility theorem regarding the relation used in ℙ𝕣𝕠𝕛. (Contributed by Steven Nguyen, 29-Apr-2023.)

Hypothesis

Ref	Expression
prjsprel.1	⊢ ∼ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵) ∧ ∃𝑙 ∈ 𝐾 𝑥 = (𝑙 · 𝑦))}

Assertion

Ref	Expression
prjsprel	⊢ (𝑋 ∼ 𝑌 ↔ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ ∃𝑚 ∈ 𝐾 𝑋 = (𝑚 · 𝑌)))

Distinct variable groups:   𝑥,𝐵,𝑦   𝑥,𝑋,𝑦,𝑙,𝑚   𝑥,𝑌,𝑦,𝑙,𝑚   𝑥,𝐾,𝑦,𝑙,𝑚   𝑥, · ,𝑦,𝑙,𝑚

Allowed substitution hints:   𝐵(𝑚, 𝑙)   ∼ (𝑥, 𝑦, 𝑚, 𝑙)

Proof of Theorem prjsprel

Step	Hyp	Ref	Expression
1		simpll 779	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑥 = 𝑋)
2		simpr 490	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑙 = 𝑚)
3		simplr 781	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑦 = 𝑌)
4	2, 3	oveq12d 7435	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑙 · 𝑦) = (𝑚 · 𝑌))
5	1, 4	eqeq12d 2778	. . 3 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑥 = (𝑙 · 𝑦) ↔ 𝑋 = (𝑚 · 𝑌)))
6	5	cbvrexdva 3245	. 2 ⊢ ((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) → (∃𝑙 ∈ 𝐾 𝑥 = (𝑙 · 𝑦) ↔ ∃𝑚 ∈ 𝐾 𝑋 = (𝑚 · 𝑌)))
7		prjsprel.1	. 2 ⊢ ∼ = {⟨𝑥, 𝑦⟩ ∣ ((𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵) ∧ ∃𝑙 ∈ 𝐾 𝑥 = (𝑙 · 𝑦))}
8	6, 7	brab2a 5752	1 ⊢ (𝑋 ∼ 𝑌 ↔ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ ∃𝑚 ∈ 𝐾 𝑋 = (𝑚 · 𝑌)))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∃wrex 3088   class class class wbr 5107  {copab 5171  (class class class)co 7417

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 2734  ax-sep 5255  ax-pr 5402

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 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-xp 5665  df-iota 6493  df-fv 6545  df-ov 7420

This theorem is used by:  prjspertr  43459  prjsperref  43460  prjspersym  43461  prjspreln0  43463  prjspvs  43464  prjspner1  43480  0prjspnrel  43481