Theorem p0val 18382

Description: Value of poset zero. (Contributed by NM, 12-Oct-2011.)

Hypotheses

Ref	Expression
p0val.b	⊢ 𝐵 = (Base‘𝐾)
p0val.g	⊢ 𝐺 = (glb‘𝐾)
p0val.z	⊢ 0 = (0.‘𝐾)

Assertion

Ref	Expression
p0val	⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Proof of Theorem p0val

Dummy variable 𝑝 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elex 3451	. 2 ⊢ (𝐾 ∈ 𝑉 → 𝐾 ∈ V)
2		p0val.z	. . 3 ⊢ 0 = (0.‘𝐾)
3		fveq2 6834	. . . . . 6 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = (glb‘𝐾))
4		p0val.g	. . . . . 6 ⊢ 𝐺 = (glb‘𝐾)
5	3, 4	eqtr4di 2790	. . . . 5 ⊢ (𝑝 = 𝐾 → (glb‘𝑝) = 𝐺)
6		fveq2 6834	. . . . . 6 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = (Base‘𝐾))
7		p0val.b	. . . . . 6 ⊢ 𝐵 = (Base‘𝐾)
8	6, 7	eqtr4di 2790	. . . . 5 ⊢ (𝑝 = 𝐾 → (Base‘𝑝) = 𝐵)
9	5, 8	fveq12d 6841	. . . 4 ⊢ (𝑝 = 𝐾 → ((glb‘𝑝)‘(Base‘𝑝)) = (𝐺‘𝐵))
10		df-p0 18380	. . . 4 ⊢ 0. = (𝑝 ∈ V ↦ ((glb‘𝑝)‘(Base‘𝑝)))
11		fvex 6847	. . . 4 ⊢ (𝐺‘𝐵) ∈ V
12	9, 10, 11	fvmpt 6941	. . 3 ⊢ (𝐾 ∈ V → (0.‘𝐾) = (𝐺‘𝐵))
13	2, 12	eqtrid 2784	. 2 ⊢ (𝐾 ∈ V → 0 = (𝐺‘𝐵))
14	1, 13	syl 17	1 ⊢ (𝐾 ∈ 𝑉 → 0 = (𝐺‘𝐵))

Syntax hints:   → wi 4   = wceq 1542   ∈ wcel 2114  Vcvv 3430  ‘cfv 6492  Basecbs 17170  glbcglb 18267  0.cp0 18378

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-10 2147  ax-11 2163  ax-12 2185  ax-ext 2709  ax-sep 5231  ax-nul 5241  ax-pr 5370

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-nf 1786  df-sb 2069  df-mo 2540  df-eu 2570  df-clab 2716  df-cleq 2729  df-clel 2812  df-nfc 2886  df-ne 2934  df-ral 3053  df-rex 3063  df-rab 3391  df-v 3432  df-dif 3893  df-un 3895  df-in 3897  df-ss 3907  df-nul 4275  df-if 4468  df-sn 4569  df-pr 4571  df-op 4575  df-uni 4852  df-br 5087  df-opab 5149  df-mpt 5168  df-id 5519  df-xp 5630  df-rel 5631  df-cnv 5632  df-co 5633  df-dm 5634  df-iota 6448  df-fun 6494  df-fv 6500  df-p0 18380

This theorem is referenced by:  p0le  18384  clatp0cl  33051  xrsp0  33087  op0cl  39644  atl0cl  39763