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Theorem p0val 18579
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 3472 . 2 (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2 p0val.z . . 3 0 = (0.‘𝐾)
3 fveq2 6877 . . . . . 6 (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4 p0val.g . . . . . 6 𝐺 = (glb‘𝐾)
53, 4eqtr4di 2814 . . . . 5 (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6 fveq2 6877 . . . . . 6 (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7 p0val.b . . . . . 6 𝐵 = (Base‘𝐾)
86, 7eqtr4di 2814 . . . . 5 (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
95, 8fveq12d 6884 . . . 4 (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10 df-p0 18577 . . . 4 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11 fvex 6890 . . . 4 (𝐺‘𝐵) ∈ V
129, 10, 11fvmpt 6985 . . 3 (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
132, 12eqtrid 2808 . 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 1570   ∈ wcel 2145  Vcvv 3451  ‘cfv 6531  Basecbs 17367  glbcglb 18464  0.cp0 18575
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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-nul 5260  ax-pr 5391
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 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-iota 6487  df-fun 6533  df-fv 6539  df-p0 18577
This theorem is used by:  p0le  18581  clatp0cl  33519  xrsp0  33555  op0cl  40209  atl0cl  40328
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