Theorem p1val 16865

Description: Value of poset zero. (Contributed by NM, 22-Oct-2011.)

Hypotheses

Ref	Expression
p1val.b	⊢ 𝐵 = (Base‘𝐾)
p1val.u	⊢ 𝑈 = (lub‘𝐾)
p1val.t	⊢ 1 = (1.‘𝐾)

Assertion

Ref	Expression
p1val	⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Proof of Theorem p1val

Dummy variable 𝑘 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elex 3185	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6103	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	syl6eqr 2662	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6103	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	syl6eqr 2662	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6109	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 16863	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6113	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6191	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	syl5eq 2656	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1475   ∈ wcel 1977  Vcvv 3173  ‘cfv 5804  Basecbs 15695  lubclub 16765  1.cp1 16861

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pr 4833

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-sbc 3403  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-uni 4373  df-br 4584  df-opab 4644  df-mpt 4645  df-id 4953  df-xp 5044  df-rel 5045  df-cnv 5046  df-co 5047  df-dm 5048  df-iota 5768  df-fun 5806  df-fv 5812  df-p1 16863

This theorem is referenced by:  ple1  16867  clatp1cl  29003  xrsp1  29013  op1cl  33490