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Theorem p0val 18605
Description: Value of poset zero. (Contributed by NM, 12-Oct-2011.)
Hypotheses
Ref Expression
p0val.b 𝐵 = (Base‘𝐾)
p0val.g 𝐺 = (glb‘𝐾)
p0val.z 0 = (0.‘𝐾)
Assertion
Ref Expression
p0val (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Proof of Theorem p0val
Dummy variable 𝑝 is distinct from all other variables.
StepHypRef Expression
1 elex 3471 . 2 (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2 p0val.z . . 3 0 = (0.‘𝐾)
3 fveq2 6886 . . . . . 6 (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4 p0val.g . . . . . 6 𝐺 = (glb‘𝐾)
53, 4eqtr4di 2813 . . . . 5 (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6 fveq2 6886 . . . . . 6 (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7 p0val.b . . . . . 6 𝐵 = (Base‘𝐾)
86, 7eqtr4di 2813 . . . . 5 (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
95, 8fveq12d 6893 . . . 4 (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10 df-p0 18603 . . . 4 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11 fvex 6899 . . . 4 (𝐺‘𝐵) ∈ V
129, 10, 11fvmpt 6994 . . 3 (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
132, 12eqtrid 2807 . 2 (𝐾 ∈ V → 0 = (𝐺‘𝐵))
141, 13syl 18 1 (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∈ wcel 2144  Vcvv 3450  ‘cfv 6539  Basecbs 17393  glbcglb 18490  0.cp0 18601
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pr 5390
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-uni 4867  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-iota 6494  df-fun 6541  df-fv 6547  df-p0 18603
This theorem is used by:  p0le  18607  clatp0cl  33584  xrsp0  33620  op0cl  40335  atl0cl  40454
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