Theorem p1val 17652

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 3512	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6670	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	syl6eqr 2874	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6670	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	syl6eqr 2874	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6677	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 17650	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6683	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6768	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	syl5eq 2868	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1537   ∈ wcel 2114  Vcvv 3494  ‘cfv 6355  Basecbs 16483  lubclub 17552  1.cp1 17648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793  ax-sep 5203  ax-nul 5210  ax-pr 5330

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-mo 2622  df-eu 2654  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rex 3144  df-rab 3147  df-v 3496  df-sbc 3773  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4568  df-pr 4570  df-op 4574  df-uni 4839  df-br 5067  df-opab 5129  df-mpt 5147  df-id 5460  df-xp 5561  df-rel 5562  df-cnv 5563  df-co 5564  df-dm 5565  df-iota 6314  df-fun 6357  df-fv 6363  df-p1 17650

This theorem is referenced by:  ple1  17654  clatp1cl  30659  xrsp1  30669  op1cl  36336