Theorem p0val 18442

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 3485	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6881	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2789	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6881	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2789	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6888	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 18440	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6894	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6991	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	eqtrid 2783	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1540   ∈ wcel 2109  Vcvv 3464  ‘cfv 6536  Basecbs 17233  glbcglb 18327  0.cp0 18438

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2708  ax-sep 5271  ax-nul 5281  ax-pr 5407

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2728  df-clel 2810  df-nfc 2886  df-ne 2934  df-ral 3053  df-rex 3062  df-rab 3421  df-v 3466  df-dif 3934  df-un 3936  df-ss 3948  df-nul 4314  df-if 4506  df-sn 4607  df-pr 4609  df-op 4613  df-uni 4889  df-br 5125  df-opab 5187  df-mpt 5207  df-id 5553  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-iota 6489  df-fun 6538  df-fv 6544  df-p0 18440

This theorem is referenced by:  p0le  18444  clatp0cl  32961  xrsp0  33009  op0cl  39207  atl0cl  39326