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Theorem p0val 18506
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 (𝐾𝑉0 = (𝐺𝐵))

Proof of Theorem p0val
Dummy variable 𝑝 is distinct from all other variables.
StepHypRef Expression
1 elex 3479 . 2 (𝐾𝑉𝐾 ∈ V)
2 p0val.z . . 3 0 = (0.‘𝐾)
3 fveq2 6888 . . . . . 6 (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4 p0val.g . . . . . 6 𝐺 = (glb‘𝐾)
53, 4eqtr4di 2819 . . . . 5 (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6 fveq2 6888 . . . . . 6 (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7 p0val.b . . . . . 6 𝐵 = (Base‘𝐾)
86, 7eqtr4di 2819 . . . . 5 (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
95, 8fveq12d 6895 . . . 4 (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺𝐵))
10 df-p0 18504 . . . 4 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11 fvex 6901 . . . 4 (𝐺𝐵) ∈ V
129, 10, 11fvmpt 6996 . . 3 (𝐾 ∈ V → (0.‘𝐾) = (𝐺𝐵))
132, 12eqtrid 2813 . 2 (𝐾 ∈ V → 0 = (𝐺𝐵))
141, 13syl 18 1 (𝐾𝑉0 = (𝐺𝐵))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wi 4   = wceq 1570  wcel 2146  Vcvv 3458  cfv 6543  Basecbs 17294  glbcglb 18391  0.cp0 18502
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738  ax-sep 5262  ax-nul 5274  ax-pr 5409
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2570  df-eu 2600  df-clab 2745  df-cleq 2758  df-clel 2841  df-nfc 2915  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-opab 5179  df-mpt 5198  df-id 5561  df-xp 5672  df-rel 5673  df-cnv 5674  df-co 5675  df-dm 5676  df-iota 6499  df-fun 6545  df-fv 6551  df-p0 18504
This theorem is used by:  p0le  18508  clatp0cl  33327  xrsp0  33363  op0cl  39999  atl0cl  40118
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