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Theorem recexprlemex 6735
 Description: 𝐵 is the reciprocal of 𝐴. Lemma for recexpr 6736. (Contributed by Jim Kingdon, 27-Dec-2019.)
Hypothesis
Ref Expression
recexpr.1 𝐵 = ⟨{𝑥 ∣ ∃𝑦(𝑥 <Q 𝑦 ∧ (*Q𝑦) ∈ (2nd𝐴))}, {𝑥 ∣ ∃𝑦(𝑦 <Q 𝑥 ∧ (*Q𝑦) ∈ (1st𝐴))}⟩
Assertion
Ref Expression
recexprlemex (𝐴P → (𝐴 ·P 𝐵) = 1P)
Distinct variable groups:   𝑥,𝑦,𝐴   𝑥,𝐵,𝑦

Proof of Theorem recexprlemex
StepHypRef Expression
1 recexpr.1 . . . 4 𝐵 = ⟨{𝑥 ∣ ∃𝑦(𝑥 <Q 𝑦 ∧ (*Q𝑦) ∈ (2nd𝐴))}, {𝑥 ∣ ∃𝑦(𝑦 <Q 𝑥 ∧ (*Q𝑦) ∈ (1st𝐴))}⟩
21recexprlemss1l 6733 . . 3 (𝐴P → (1st ‘(𝐴 ·P 𝐵)) ⊆ (1st ‘1P))
31recexprlem1ssl 6731 . . 3 (𝐴P → (1st ‘1P) ⊆ (1st ‘(𝐴 ·P 𝐵)))
42, 3eqssd 2962 . 2 (𝐴P → (1st ‘(𝐴 ·P 𝐵)) = (1st ‘1P))
51recexprlemss1u 6734 . . 3 (𝐴P → (2nd ‘(𝐴 ·P 𝐵)) ⊆ (2nd ‘1P))
61recexprlem1ssu 6732 . . 3 (𝐴P → (2nd ‘1P) ⊆ (2nd ‘(𝐴 ·P 𝐵)))
75, 6eqssd 2962 . 2 (𝐴P → (2nd ‘(𝐴 ·P 𝐵)) = (2nd ‘1P))
81recexprlempr 6730 . . . 4 (𝐴P𝐵P)
9 mulclpr 6670 . . . 4 ((𝐴P𝐵P) → (𝐴 ·P 𝐵) ∈ P)
108, 9mpdan 398 . . 3 (𝐴P → (𝐴 ·P 𝐵) ∈ P)
11 1pr 6652 . . 3 1PP
12 preqlu 6570 . . 3 (((𝐴 ·P 𝐵) ∈ P ∧ 1PP) → ((𝐴 ·P 𝐵) = 1P ↔ ((1st ‘(𝐴 ·P 𝐵)) = (1st ‘1P) ∧ (2nd ‘(𝐴 ·P 𝐵)) = (2nd ‘1P))))
1310, 11, 12sylancl 392 . 2 (𝐴P → ((𝐴 ·P 𝐵) = 1P ↔ ((1st ‘(𝐴 ·P 𝐵)) = (1st ‘1P) ∧ (2nd ‘(𝐴 ·P 𝐵)) = (2nd ‘1P))))
144, 7, 13mpbir2and 851 1 (𝐴P → (𝐴 ·P 𝐵) = 1P)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98   = wceq 1243  ∃wex 1381   ∈ wcel 1393  {cab 2026  ⟨cop 3378   class class class wbr 3764  ‘cfv 4902  (class class class)co 5512  1st c1st 5765  2nd c2nd 5766  *Qcrq 6382
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