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 3468	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p1val.t	. . 3 ⊢ 1 = (1.‘𝐾)
3		fveq2 6858	. . . . . 6 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = (lub‘𝐾))
4		p1val.u	. . . . . 6 ⊢ 𝑈 = (lub‘𝐾)
5	3, 4	eqtr4di 2782	. . . . 5 ⊢ (𝑘 = 𝐾 → (lub‘𝑘) = 𝑈)
6		fveq2 6858	. . . . . 6 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = (Base‘𝐾))
7		p1val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2782	. . . . 5 ⊢ (𝑘 = 𝐾 → (Base‘𝑘) = 𝐵)
9	5, 8	fveq12d 6865	. . . 4 ⊢ (𝑘 = 𝐾 → ((lub‘𝑘)‘(Base‘𝑘)) = (𝑈‘𝐵))
10		df-p1 18385	. . . 4 ⊢ 1. = (𝑘 ∈ V ↦ ((lub‘𝑘)‘(Base‘𝑘)))
11		fvex 6871	. . . 4 ⊢ (𝑈‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6968	. . 3 ⊢ (𝐾 ∈ V → (1.‘𝐾) = (𝑈‘𝐵))
13	2, 12	eqtrid 2776	. 2 ⊢ (𝐾 ∈ V → 1 = (𝑈‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 1 = (𝑈‘𝐵))

Syntax hints:   → wi 4   = wceq 1540   ∈ wcel 2109  Vcvv 3447  ‘cfv 6511  Basecbs 17179  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 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-10 2142  ax-11 2158  ax-12 2178  ax-ext 2701  ax-sep 5251  ax-nul 5261  ax-pr 5387

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2066  df-mo 2533  df-eu 2562  df-clab 2708  df-cleq 2721  df-clel 2803  df-nfc 2878  df-ne 2926  df-ral 3045  df-rex 3054  df-rab 3406  df-v 3449  df-dif 3917  df-un 3919  df-ss 3931  df-nul 4297  df-if 4489  df-sn 4590  df-pr 4592  df-op 4596  df-uni 4872  df-br 5108  df-opab 5170  df-mpt 5189  df-id 5533  df-xp 5644  df-rel 5645  df-cnv 5646  df-co 5647  df-dm 5648  df-iota 6464  df-fun 6513  df-fv 6519  df-p1 18385

This theorem is referenced by:  ple1  18389  clatp1cl  32903  xrsp1  32951  op1cl  39178