Theorem p1val 18387

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 3454	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6830	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	eqtr4di 2794	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6830	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2794	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6837	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 18385	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6843	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6938	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	eqtrid 2788	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1548   ∈ wcel 2121  Vcvv 3433  ‘cfv 6488  Basecbs 17174  lubclub 18270  1.cp1 18383

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1975  ax-7 2016  ax-8 2123  ax-9 2131  ax-10 2154  ax-11 2170  ax-12 2191  ax-ext 2713  ax-sep 5220  ax-nul 5230  ax-pr 5364

This theorem depends on definitions:  df-bi 209  df-an 398  df-or 855  df-3an 1095  df-tru 1551  df-fal 1561  df-ex 1788  df-nf 1792  df-sb 2075  df-mo 2545  df-eu 2575  df-clab 2720  df-cleq 2733  df-clel 2816  df-nfc 2890  df-ne 2937  df-ral 3056  df-rex 3066  df-rab 3394  df-v 3435  df-dif 3887  df-un 3889  df-in 3891  df-ss 3901  df-nul 4264  df-if 4457  df-sn 4558  df-pr 4560  df-op 4564  df-uni 4841  df-br 5075  df-opab 5137  df-mpt 5156  df-id 5515  df-xp 5626  df-rel 5627  df-cnv 5628  df-co 5629  df-dm 5630  df-iota 6444  df-fun 6490  df-fv 6496  df-p1 18385

This theorem is referenced by:  ple1  18389  clatp1cl  33058  xrsp1  33094  op1cl  39690