Theorem prjsprel 42493

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 766	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑥 = 𝑋)
2		simpr 484	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑙 = 𝑚)
3		simplr 768	. . . . 5 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → 𝑦 = 𝑌)
4	2, 3	oveq12d 7463	. . . 4 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑙 · 𝑦) = (𝑚 · 𝑌))
5	1, 4	eqeq12d 2750	. . 3 ⊢ (((𝑥 = 𝑋 ∧ 𝑦 = 𝑌) ∧ 𝑙 = 𝑚) → (𝑥 = (𝑙 · 𝑦) ↔ 𝑋 = (𝑚 · 𝑌)))
6	5	cbvrexdva 3241	. 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 5792	1 ⊢ (𝑋 ∼ 𝑌 ↔ ((𝑋 ∈ 𝐵 ∧ 𝑌 ∈ 𝐵) ∧ ∃𝑚 ∈ 𝐾 𝑋 = (𝑚 · 𝑌)))

Syntax hints:   ↔ wb 206   ∧ wa 395   = wceq 1537   ∈ wcel 2103  ∃wrex 3072   class class class wbr 5169  {copab 5231  (class class class)co 7445

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2105  ax-9 2113  ax-ext 2705  ax-sep 5320  ax-nul 5327  ax-pr 5450

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-sb 2065  df-clab 2712  df-cleq 2726  df-clel 2813  df-ral 3064  df-rex 3073  df-rab 3439  df-v 3484  df-dif 3973  df-un 3975  df-ss 3987  df-nul 4348  df-if 4549  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-br 5170  df-opab 5232  df-xp 5705  df-iota 6524  df-fv 6580  df-ov 7448

This theorem is referenced by:  prjspertr  42494  prjsperref  42495  prjspersym  42496  prjspreln0  42498  prjspvs  42499  prjspner1  42515  0prjspnrel  42516