Theorem p0val 18418

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 3490	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6897	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2786	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6897	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2786	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6904	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 18416	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6910	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 7005	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	eqtrid 2780	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1534   ∈ wcel 2099  Vcvv 3471  ‘cfv 6548  Basecbs 17179  glbcglb 18301  0.cp0 18414

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-8 2101  ax-9 2109  ax-10 2130  ax-11 2147  ax-12 2167  ax-ext 2699  ax-sep 5299  ax-nul 5306  ax-pr 5429

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-3an 1087  df-tru 1537  df-fal 1547  df-ex 1775  df-nf 1779  df-sb 2061  df-mo 2530  df-eu 2559  df-clab 2706  df-cleq 2720  df-clel 2806  df-nfc 2881  df-ne 2938  df-ral 3059  df-rex 3068  df-rab 3430  df-v 3473  df-dif 3950  df-un 3952  df-in 3954  df-ss 3964  df-nul 4324  df-if 4530  df-sn 4630  df-pr 4632  df-op 4636  df-uni 4909  df-br 5149  df-opab 5211  df-mpt 5232  df-id 5576  df-xp 5684  df-rel 5685  df-cnv 5686  df-co 5687  df-dm 5688  df-iota 6500  df-fun 6550  df-fv 6556  df-p0 18416

This theorem is referenced by:  p0le  18420  clatp0cl  32703  xrsp0  32739  op0cl  38656  atl0cl  38775