Theorem p0val 18243

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 3460	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6830	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2795	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6830	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2795	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6837	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 18241	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6843	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6936	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	eqtrid 2789	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1541   ∈ wcel 2106  Vcvv 3442  ‘cfv 6484  Basecbs 17010  glbcglb 18126  0.cp0 18239

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2708  ax-sep 5248  ax-nul 5255  ax-pr 5377

This theorem depends on definitions:  df-bi 206  df-an 398  df-or 846  df-3an 1089  df-tru 1544  df-fal 1554  df-ex 1782  df-nf 1786  df-sb 2068  df-mo 2539  df-eu 2568  df-clab 2715  df-cleq 2729  df-clel 2815  df-nfc 2887  df-ne 2942  df-ral 3063  df-rex 3072  df-rab 3405  df-v 3444  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4275  df-if 4479  df-sn 4579  df-pr 4581  df-op 4585  df-uni 4858  df-br 5098  df-opab 5160  df-mpt 5181  df-id 5523  df-xp 5631  df-rel 5632  df-cnv 5633  df-co 5634  df-dm 5635  df-iota 6436  df-fun 6486  df-fv 6492  df-p0 18241

This theorem is referenced by:  p0le  18245  clatp0cl  31539  xrsp0  31575  op0cl  37500  atl0cl  37619