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Theorem hoidifhspval2 40829
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:   𝑥,𝑘   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 40822 . 2 (𝜑 → (𝐷𝑌) = (𝑎 ∈ (ℝ ↑𝑚 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)))))
4 fveq1 6190 . . . . . . 7 (𝑎 = 𝐴 → (𝑎𝑘) = (𝐴𝑘))
54breq2d 4665 . . . . . 6 (𝑎 = 𝐴 → (𝑌 ≤ (𝑎𝑘) ↔ 𝑌 ≤ (𝐴𝑘)))
65, 4ifbieq1d 4109 . . . . 5 (𝑎 = 𝐴 → if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌) = if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌))
76, 4ifeq12d 4106 . . . 4 (𝑎 = 𝐴 → if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)) = if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘)))
87mpteq2dv 4745 . . 3 (𝑎 = 𝐴 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘))) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
98adantl 482 . 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 10027 . . . . . 6 ℝ ∈ V
1211a1i 11 . . . . 5 (𝜑 → ℝ ∈ V)
13 hoidifhspval2.x . . . . 5 (𝜑𝑋𝑉)
1412, 13jca 554 . . . 4 (𝜑 → (ℝ ∈ V ∧ 𝑋𝑉))
15 elmapg 7870 . . . 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 247 . 2 (𝜑𝐴 ∈ (ℝ ↑𝑚 𝑋))
18 mptexg 6484 . . 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 6288 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 196  wa 384   = wceq 1483  wcel 1990  Vcvv 3200  ifcif 4086   class class class wbr 4653  cmpt 4729  wf 5884  cfv 5888  (class class class)co 6650  𝑚 cmap 7857  cr 9935  cle 10075
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-8 1992  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-rep 4771  ax-sep 4781  ax-nul 4789  ax-pow 4843  ax-pr 4906  ax-un 6949  ax-cnex 9992  ax-resscn 9993
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-ral 2917  df-rex 2918  df-reu 2919  df-rab 2921  df-v 3202  df-sbc 3436  df-csb 3534  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-pw 4160  df-sn 4178  df-pr 4180  df-op 4184  df-uni 4437  df-iun 4522  df-br 4654  df-opab 4713  df-mpt 4730  df-id 5024  df-xp 5120  df-rel 5121  df-cnv 5122  df-co 5123  df-dm 5124  df-rn 5125  df-res 5126  df-ima 5127  df-iota 5851  df-fun 5890  df-fn 5891  df-f 5892  df-f1 5893  df-fo 5894  df-f1o 5895  df-fv 5896  df-ov 6653  df-oprab 6654  df-mpt2 6655  df-map 7859
This theorem is referenced by:  hoidifhspf  40832  hoidifhspval3  40833
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