Theorem p1val 17655

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 3515	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6673	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	syl6eqr 2877	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6673	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	syl6eqr 2877	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6680	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 17653	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6686	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6771	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	syl5eq 2871	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1536   ∈ wcel 2113  Vcvv 3497  ‘cfv 6358  Basecbs 16486  lubclub 17555  1.cp1 17651

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 1969  ax-7 2014  ax-8 2115  ax-9 2123  ax-10 2144  ax-11 2160  ax-12 2176  ax-ext 2796  ax-sep 5206  ax-nul 5213  ax-pr 5333

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1539  df-ex 1780  df-nf 1784  df-sb 2069  df-mo 2621  df-eu 2653  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2966  df-ral 3146  df-rex 3147  df-rab 3150  df-v 3499  df-sbc 3776  df-dif 3942  df-un 3944  df-in 3946  df-ss 3955  df-nul 4295  df-if 4471  df-sn 4571  df-pr 4573  df-op 4577  df-uni 4842  df-br 5070  df-opab 5132  df-mpt 5150  df-id 5463  df-xp 5564  df-rel 5565  df-cnv 5566  df-co 5567  df-dm 5568  df-iota 6317  df-fun 6360  df-fv 6366  df-p1 17653

This theorem is referenced by:  ple1  17657  clatp1cl  30663  xrsp1  30673  op1cl  36325