Theorem prjsprel 43150

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 776	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑥 = 𝑋)
2		simpr 488	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑙 = 𝑚)
3		simplr 778	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑦 = 𝑌)
4	2, 3	oveq12d 7410	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑙 · 𝑦) = (𝑚 · 𝑌))
5	1, 4	eqeq12d 2777	. . 3 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑥 = (𝑙 · 𝑦) ↔ 𝑋 = (𝑚 · 𝑌)))
6	5	cbvrexdva 3242	. 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 5738	1 ⊢ (𝑋 ∼ 𝑌 ↔ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ ∃𝑚 ∈ 𝐾 𝑋 = (𝑚 · 𝑌)))

Syntax hints:   ↔ wb 208   ∧ wa 399   = wceq 1559   ∈ wcel 2141  ∃wrex 3085   class class class wbr 5099  {copab 5161  (class class class)co 7392

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733  ax-sep 5245  ax-pr 5389

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-br 5100  df-opab 5162  df-xp 5651  df-iota 6473  df-fv 6525  df-ov 7395

This theorem is referenced by:  prjspertr  43151  prjsperref  43152  prjspersym  43153  prjspreln0  43155  prjspvs  43156  prjspner1  43172  0prjspnrel  43173