Theorem p0val 17643

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 3459	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6645	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2851	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6645	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2851	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6652	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 17641	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6658	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6745	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	syl5eq 2845	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1538   ∈ wcel 2111  Vcvv 3441  ‘cfv 6324  Basecbs 16475  glbcglb 17545  0.cp0 17639

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 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2770  ax-sep 5167  ax-nul 5174  ax-pr 5295

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3an 1086  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-mo 2598  df-eu 2629  df-clab 2777  df-cleq 2791  df-clel 2870  df-nfc 2938  df-ral 3111  df-rex 3112  df-v 3443  df-sbc 3721  df-dif 3884  df-un 3886  df-in 3888  df-ss 3898  df-nul 4244  df-if 4426  df-sn 4526  df-pr 4528  df-op 4532  df-uni 4801  df-br 5031  df-opab 5093  df-mpt 5111  df-id 5425  df-xp 5525  df-rel 5526  df-cnv 5527  df-co 5528  df-dm 5529  df-iota 6283  df-fun 6326  df-fv 6332  df-p0 17641

This theorem is referenced by:  p0le  17645  clatp0cl  30684  xrsp0  30715  op0cl  36480  atl0cl  36599