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Theorem genpcl 10423
 Description: Closure of an operation on reals. (Contributed by NM, 13-Mar-1996.) (Revised by Mario Carneiro, 17-Nov-2014.) (New usage is discouraged.)
Hypotheses
Ref Expression
genp.1 𝐹 = (𝑤P, 𝑣P ↦ {𝑥 ∣ ∃𝑦𝑤𝑧𝑣 𝑥 = (𝑦𝐺𝑧)})
genp.2 ((𝑦Q𝑧Q) → (𝑦𝐺𝑧) ∈ Q)
genpcl.3 (Q → (𝑓 <Q 𝑔 ↔ (𝐺𝑓) <Q (𝐺𝑔)))
genpcl.4 (𝑥𝐺𝑦) = (𝑦𝐺𝑥)
genpcl.5 ((((𝐴P𝑔𝐴) ∧ (𝐵P𝐵)) ∧ 𝑥Q) → (𝑥 <Q (𝑔𝐺) → 𝑥 ∈ (𝐴𝐹𝐵)))
Assertion
Ref Expression
genpcl ((𝐴P𝐵P) → (𝐴𝐹𝐵) ∈ P)
Distinct variable groups:   𝑥,𝑦,𝑧,𝑓,𝑔,,𝐴   𝑥,𝐵,𝑦,𝑧,𝑓,𝑔,,𝑤,𝑣   𝑥,𝐺   𝑦,𝑤,𝑣,𝐺,𝑧,𝑓,𝑔,   𝑓,𝐹,𝑔   𝑤,𝐴,𝑣   𝑤,𝐵,𝑣   𝑥,𝐹,𝑦,𝑤,𝑣,
Allowed substitution hint:   𝐹(𝑧)

Proof of Theorem genpcl
StepHypRef Expression
1 genp.1 . . 3 𝐹 = (𝑤P, 𝑣P ↦ {𝑥 ∣ ∃𝑦𝑤𝑧𝑣 𝑥 = (𝑦𝐺𝑧)})
2 genp.2 . . 3 ((𝑦Q𝑧Q) → (𝑦𝐺𝑧) ∈ Q)
31, 2genpn0 10418 . 2 ((𝐴P𝐵P) → ∅ ⊊ (𝐴𝐹𝐵))
41, 2genpss 10419 . . 3 ((𝐴P𝐵P) → (𝐴𝐹𝐵) ⊆ Q)
5 vex 3447 . . . . 5 𝑥 ∈ V
6 vex 3447 . . . . 5 𝑦 ∈ V
7 genpcl.3 . . . . 5 (Q → (𝑓 <Q 𝑔 ↔ (𝐺𝑓) <Q (𝐺𝑔)))
85, 6, 7caovord 7343 . . . 4 (𝑧Q → (𝑥 <Q 𝑦 ↔ (𝑧𝐺𝑥) <Q (𝑧𝐺𝑦)))
9 genpcl.4 . . . 4 (𝑥𝐺𝑦) = (𝑦𝐺𝑥)
101, 2, 8, 9genpnnp 10420 . . 3 ((𝐴P𝐵P) → ¬ (𝐴𝐹𝐵) = Q)
11 dfpss2 4016 . . 3 ((𝐴𝐹𝐵) ⊊ Q ↔ ((𝐴𝐹𝐵) ⊆ Q ∧ ¬ (𝐴𝐹𝐵) = Q))
124, 10, 11sylanbrc 586 . 2 ((𝐴P𝐵P) → (𝐴𝐹𝐵) ⊊ Q)
13 genpcl.5 . . . . . 6 ((((𝐴P𝑔𝐴) ∧ (𝐵P𝐵)) ∧ 𝑥Q) → (𝑥 <Q (𝑔𝐺) → 𝑥 ∈ (𝐴𝐹𝐵)))
141, 2, 13genpcd 10421 . . . . 5 ((𝐴P𝐵P) → (𝑓 ∈ (𝐴𝐹𝐵) → (𝑥 <Q 𝑓𝑥 ∈ (𝐴𝐹𝐵))))
1514alrimdv 1930 . . . 4 ((𝐴P𝐵P) → (𝑓 ∈ (𝐴𝐹𝐵) → ∀𝑥(𝑥 <Q 𝑓𝑥 ∈ (𝐴𝐹𝐵))))
16 vex 3447 . . . . . 6 𝑧 ∈ V
17 vex 3447 . . . . . 6 𝑤 ∈ V
1816, 17, 7caovord 7343 . . . . 5 (𝑣Q → (𝑧 <Q 𝑤 ↔ (𝑣𝐺𝑧) <Q (𝑣𝐺𝑤)))
1916, 17, 9caovcom 7329 . . . . 5 (𝑧𝐺𝑤) = (𝑤𝐺𝑧)
201, 2, 18, 19genpnmax 10422 . . . 4 ((𝐴P𝐵P) → (𝑓 ∈ (𝐴𝐹𝐵) → ∃𝑥 ∈ (𝐴𝐹𝐵)𝑓 <Q 𝑥))
2115, 20jcad 516 . . 3 ((𝐴P𝐵P) → (𝑓 ∈ (𝐴𝐹𝐵) → (∀𝑥(𝑥 <Q 𝑓𝑥 ∈ (𝐴𝐹𝐵)) ∧ ∃𝑥 ∈ (𝐴𝐹𝐵)𝑓 <Q 𝑥)))
2221ralrimiv 3151 . 2 ((𝐴P𝐵P) → ∀𝑓 ∈ (𝐴𝐹𝐵)(∀𝑥(𝑥 <Q 𝑓𝑥 ∈ (𝐴𝐹𝐵)) ∧ ∃𝑥 ∈ (𝐴𝐹𝐵)𝑓 <Q 𝑥))
23 elnp 10402 . 2 ((𝐴𝐹𝐵) ∈ P ↔ ((∅ ⊊ (𝐴𝐹𝐵) ∧ (𝐴𝐹𝐵) ⊊ Q) ∧ ∀𝑓 ∈ (𝐴𝐹𝐵)(∀𝑥(𝑥 <Q 𝑓𝑥 ∈ (𝐴𝐹𝐵)) ∧ ∃𝑥 ∈ (𝐴𝐹𝐵)𝑓 <Q 𝑥)))
243, 12, 22, 23syl21anbrc 1341 1 ((𝐴P𝐵P) → (𝐴𝐹𝐵) ∈ P)
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536   = wceq 1538   ∈ wcel 2112  {cab 2779  ∀wral 3109  ∃wrex 3110   ⊆ wss 3884   ⊊ wpss 3885  ∅c0 4246   class class class wbr 5033  (class class class)co 7139   ∈ cmpo 7141  Qcnq 10267
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