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Theorem hoidifhspval2 46141
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 46134 . 2 (𝜑 → (𝐷𝑌) = (𝑎 ∈ (ℝ ↑m 𝑋) ↦ (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)))))
4 fveq1 6895 . . . . . . 7 (𝑎 = 𝐴 → (𝑎𝑘) = (𝐴𝑘))
54breq2d 5161 . . . . . 6 (𝑎 = 𝐴 → (𝑌 ≤ (𝑎𝑘) ↔ 𝑌 ≤ (𝐴𝑘)))
65, 4ifbieq1d 4554 . . . . 5 (𝑎 = 𝐴 → if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌) = if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌))
76, 4ifeq12d 4551 . . . 4 (𝑎 = 𝐴 → if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘)) = if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘)))
87mpteq2dv 5251 . . 3 (𝑎 = 𝐴 → (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝑎𝑘), (𝑎𝑘), 𝑌), (𝑎𝑘))) = (𝑘𝑋 ↦ if(𝑘 = 𝐾, if(𝑌 ≤ (𝐴𝑘), (𝐴𝑘), 𝑌), (𝐴𝑘))))
98adantl 480 . 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 11231 . . . . . 6 ℝ ∈ V
1211a1i 11 . . . . 5 (𝜑 → ℝ ∈ V)
13 hoidifhspval2.x . . . . 5 (𝜑𝑋𝑉)
1412, 13jca 510 . . . 4 (𝜑 → (ℝ ∈ V ∧ 𝑋𝑉))
15 elmapg 8858 . . . 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 256 . 2 (𝜑𝐴 ∈ (ℝ ↑m 𝑋))
18 mptexg 7233 . . 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 7011 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 205  wa 394   = wceq 1533  wcel 2098  Vcvv 3461  ifcif 4530   class class class wbr 5149  cmpt 5232  wf 6545  cfv 6549  (class class class)co 7419  m cmap 8845  cr 11139  cle 11281
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-11 2146  ax-12 2166  ax-ext 2696  ax-rep 5286  ax-sep 5300  ax-nul 5307  ax-pow 5365  ax-pr 5429  ax-un 7741  ax-cnex 11196  ax-resscn 11197
This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-3an 1086  df-tru 1536  df-fal 1546  df-ex 1774  df-nf 1778  df-sb 2060  df-mo 2528  df-eu 2557  df-clab 2703  df-cleq 2717  df-clel 2802  df-nfc 2877  df-ne 2930  df-ral 3051  df-rex 3060  df-reu 3364  df-rab 3419  df-v 3463  df-sbc 3774  df-csb 3890  df-dif 3947  df-un 3949  df-in 3951  df-ss 3961  df-nul 4323  df-if 4531  df-pw 4606  df-sn 4631  df-pr 4633  df-op 4637  df-uni 4910  df-iun 4999  df-br 5150  df-opab 5212  df-mpt 5233  df-id 5576  df-xp 5684  df-rel 5685  df-cnv 5686  df-co 5687  df-dm 5688  df-rn 5689  df-res 5690  df-ima 5691  df-iota 6501  df-fun 6551  df-fn 6552  df-f 6553  df-f1 6554  df-fo 6555  df-f1o 6556  df-fv 6557  df-ov 7422  df-oprab 7423  df-mpo 7424  df-map 8847
This theorem is referenced by:  hoidifhspf  46144  hoidifhspval3  46145
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