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Theorem hoidifhspval2 42904
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 42897 . 2 (𝜑 → (𝐷𝑌) = (𝑎 ∈ (ℝ ↑m 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)))))
4 fveq1 6671 . . . . . . 7 (𝑎 = 𝐴 → (𝑎𝑘) = (𝐴𝑘))
54breq2d 5080 . . . . . 6 (𝑎 = 𝐴 → (𝑌 ≤ (𝑎𝑘) ↔ 𝑌 ≤ (𝐴𝑘)))
65, 4ifbieq1d 4492 . . . . 5 (𝑎 = 𝐴 → if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌) = if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌))
76, 4ifeq12d 4489 . . . 4 (𝑎 = 𝐴 → if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)) = if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘)))
87mpteq2dv 5164 . . 3 (𝑎 = 𝐴 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘))) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
98adantl 484 . 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 10630 . . . . . 6 ℝ ∈ V
1211a1i 11 . . . . 5 (𝜑 → ℝ ∈ V)
13 hoidifhspval2.x . . . . 5 (𝜑𝑋𝑉)
1412, 13jca 514 . . . 4 (𝜑 → (ℝ ∈ V ∧ 𝑋𝑉))
15 elmapg 8421 . . . 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 259 . 2 (𝜑𝐴 ∈ (ℝ ↑m 𝑋))
18 mptexg 6986 . . 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 6777 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 208  wa 398   = wceq 1537  wcel 2114  Vcvv 3496  ifcif 4469   class class class wbr 5068  cmpt 5148  wf 6353  cfv 6357  (class class class)co 7158  m cmap 8408  cr 10538  cle 10678
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795  ax-rep 5192  ax-sep 5205  ax-nul 5212  ax-pow 5268  ax-pr 5332  ax-un 7463  ax-cnex 10595  ax-resscn 10596
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-mo 2622  df-eu 2654  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ne 3019  df-ral 3145  df-rex 3146  df-reu 3147  df-rab 3149  df-v 3498  df-sbc 3775  df-csb 3886  df-dif 3941  df-un 3943  df-in 3945  df-ss 3954  df-nul 4294  df-if 4470  df-pw 4543  df-sn 4570  df-pr 4572  df-op 4576  df-uni 4841  df-iun 4923  df-br 5069  df-opab 5131  df-mpt 5149  df-id 5462  df-xp 5563  df-rel 5564  df-cnv 5565  df-co 5566  df-dm 5567  df-rn 5568  df-res 5569  df-ima 5570  df-iota 6316  df-fun 6359  df-fn 6360  df-f 6361  df-f1 6362  df-fo 6363  df-f1o 6364  df-fv 6365  df-ov 7161  df-oprab 7162  df-mpo 7163  df-map 8410
This theorem is referenced by:  hoidifhspf  42907  hoidifhspval3  42908
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