Theorem p0val 18060

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.

Step	Hyp	Ref	Expression
1		elex 3440	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6756	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2797	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6756	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2797	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6763	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 18058	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6769	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6857	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	eqtrid 2790	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1539   ∈ wcel 2108  Vcvv 3422  ‘cfv 6418  Basecbs 16840  glbcglb 17943  0.cp0 18056

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2156  ax-12 2173  ax-ext 2709  ax-sep 5218  ax-nul 5225  ax-pr 5347

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1784  df-nf 1788  df-sb 2069  df-mo 2540  df-eu 2569  df-clab 2716  df-cleq 2730  df-clel 2817  df-nfc 2888  df-ral 3068  df-rex 3069  df-rab 3072  df-v 3424  df-dif 3886  df-un 3888  df-in 3890  df-ss 3900  df-nul 4254  df-if 4457  df-sn 4559  df-pr 4561  df-op 4565  df-uni 4837  df-br 5071  df-opab 5133  df-mpt 5154  df-id 5480  df-xp 5586  df-rel 5587  df-cnv 5588  df-co 5589  df-dm 5590  df-iota 6376  df-fun 6420  df-fv 6426  df-p0 18058

This theorem is referenced by:  p0le  18062  clatp0cl  31156  xrsp0  31192  op0cl  37125  atl0cl  37244