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Theorem hoidifhspval2 46630
Description: 𝐷 is a function that returns the representation of the left side of the difference of a half-open interval and a half-space. Used in Lemma 115F of [Fremlin1] p. 31 . (Contributed by Glauco Siliprandi, 24-Dec-2020.)
Hypotheses
Ref Expression
hoidifhspval2.d 𝐷 = (𝑥 ∈ ℝ ↦ (𝑎 ∈ (ℝ ↑m 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑥 ≤ (𝑎𝑘), (𝑎𝑘), 𝑥), (𝑎𝑘)))))
hoidifhspval2.y (𝜑𝑌 ∈ ℝ)
hoidifhspval2.x (𝜑𝑋𝑉)
hoidifhspval2.a (𝜑𝐴:𝑋⟶ℝ)
Assertion
Ref Expression
hoidifhspval2 (𝜑 → ((𝐷𝑌)‘𝐴) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
Distinct variable groups:   𝐴,𝑎,𝑘   𝐾,𝑎,𝑥   𝑋,𝑎,𝑘,𝑥   𝑌,𝑎,𝑘,𝑥   𝜑,𝑎,𝑥
Allowed substitution hints:   𝜑(𝑘)   𝐴(𝑥)   𝐷(𝑥,𝑘,𝑎)   𝐾(𝑘)   𝑉(𝑥,𝑘,𝑎)

Proof of Theorem hoidifhspval2
StepHypRef Expression
1 hoidifhspval2.d . . 3 𝐷 = (𝑥 ∈ ℝ ↦ (𝑎 ∈ (ℝ ↑m 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑥 ≤ (𝑎𝑘), (𝑎𝑘), 𝑥), (𝑎𝑘)))))
2 hoidifhspval2.y . . 3 (𝜑𝑌 ∈ ℝ)
31, 2hoidifhspval 46623 . 2 (𝜑 → (𝐷𝑌) = (𝑎 ∈ (ℝ ↑m 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)))))
4 fveq1 6905 . . . . . . 7 (𝑎 = 𝐴 → (𝑎𝑘) = (𝐴𝑘))
54breq2d 5155 . . . . . 6 (𝑎 = 𝐴 → (𝑌 ≤ (𝑎𝑘) ↔ 𝑌 ≤ (𝐴𝑘)))
65, 4ifbieq1d 4550 . . . . 5 (𝑎 = 𝐴 → if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌) = if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌))
76, 4ifeq12d 4547 . . . 4 (𝑎 = 𝐴 → if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)) = if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘)))
87mpteq2dv 5244 . . 3 (𝑎 = 𝐴 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘))) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
98adantl 481 . 2 ((𝜑𝑎 = 𝐴) → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘))) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
10 hoidifhspval2.a . . 3 (𝜑𝐴:𝑋⟶ℝ)
11 reex 11246 . . . . . 6 ℝ ∈ V
1211a1i 11 . . . . 5 (𝜑 → ℝ ∈ V)
13 hoidifhspval2.x . . . . 5 (𝜑𝑋𝑉)
1412, 13jca 511 . . . 4 (𝜑 → (ℝ ∈ V ∧ 𝑋𝑉))
15 elmapg 8879 . . . 4 ((ℝ ∈ V ∧ 𝑋𝑉) → (𝐴 ∈ (ℝ ↑m 𝑋) ↔ 𝐴:𝑋⟶ℝ))
1614, 15syl 17 . . 3 (𝜑 → (𝐴 ∈ (ℝ ↑m 𝑋) ↔ 𝐴:𝑋⟶ℝ))
1710, 16mpbird 257 . 2 (𝜑𝐴 ∈ (ℝ ↑m 𝑋))
18 mptexg 7241 . . 3 (𝑋𝑉 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))) ∈ V)
1913, 18syl 17 . 2 (𝜑 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))) ∈ V)
203, 9, 17, 19fvmptd 7023 1 (𝜑 → ((𝐷𝑌)‘𝐴) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395   = wceq 1540  wcel 2108  Vcvv 3480  ifcif 4525   class class class wbr 5143  cmpt 5225  wf 6557  cfv 6561  (class class class)co 7431  m cmap 8866  cr 11154  cle 11296
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708  ax-rep 5279  ax-sep 5296  ax-nul 5306  ax-pow 5365  ax-pr 5432  ax-un 7755  ax-cnex 11211  ax-resscn 11212
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ne 2941  df-ral 3062  df-rex 3071  df-reu 3381  df-rab 3437  df-v 3482  df-sbc 3789  df-csb 3900  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-pw 4602  df-sn 4627  df-pr 4629  df-op 4633  df-uni 4908  df-iun 4993  df-br 5144  df-opab 5206  df-mpt 5226  df-id 5578  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694  df-dm 5695  df-rn 5696  df-res 5697  df-ima 5698  df-iota 6514  df-fun 6563  df-fn 6564  df-f 6565  df-f1 6566  df-fo 6567  df-f1o 6568  df-fv 6569  df-ov 7434  df-oprab 7435  df-mpo 7436  df-map 8868
This theorem is referenced by:  hoidifhspf  46633  hoidifhspval3  46634
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