Theorem p1val 18470

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 3478	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6871	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	eqtr4di 2818	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6871	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2818	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6878	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 18468	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6884	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6979	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	eqtrid 2812	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 18	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1563   ∈ wcel 2145  Vcvv 3457  ‘cfv 6525  Basecbs 17257  lubclub 18353  1.cp1 18466

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2215  ax-ext 2737  ax-sep 5250  ax-nul 5260  ax-pr 5394

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1103  df-tru 1566  df-fal 1576  df-ex 1803  df-nf 1807  df-sb 2094  df-mo 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ne 2961  df-ral 3080  df-rex 3090  df-rab 3418  df-v 3459  df-dif 3910  df-un 3912  df-in 3914  df-ss 3924  df-nul 4289  df-if 4484  df-sn 4586  df-pr 4588  df-op 4592  df-uni 4868  df-br 5105  df-opab 5167  df-mpt 5186  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-iota 6481  df-fun 6527  df-fv 6533  df-p1 18468

This theorem is referenced by:  ple1  18472  clatp1cl  33205  xrsp1  33241  op1cl  39816