Theorem p1val 17711

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 3429	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6659	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	eqtr4di 2812	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6659	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2812	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6666	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 17709	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6672	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6760	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	syl5eq 2806	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1539   ∈ wcel 2112  Vcvv 3410  ‘cfv 6336  Basecbs 16534  lubclub 17611  1.cp1 17707

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2114  ax-9 2122  ax-10 2143  ax-11 2159  ax-12 2176  ax-ext 2730  ax-sep 5170  ax-nul 5177  ax-pr 5299

This theorem depends on definitions:  df-bi 210  df-an 401  df-or 846  df-3an 1087  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2071  df-mo 2558  df-eu 2589  df-clab 2737  df-cleq 2751  df-clel 2831  df-nfc 2902  df-ral 3076  df-rex 3077  df-v 3412  df-sbc 3698  df-dif 3862  df-un 3864  df-in 3866  df-ss 3876  df-nul 4227  df-if 4422  df-sn 4524  df-pr 4526  df-op 4530  df-uni 4800  df-br 5034  df-opab 5096  df-mpt 5114  df-id 5431  df-xp 5531  df-rel 5532  df-cnv 5533  df-co 5534  df-dm 5535  df-iota 6295  df-fun 6338  df-fv 6344  df-p1 17709

This theorem is referenced by:  ple1  17713  clatp1cl  30774  xrsp1  30810  op1cl  36754