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Theorem sltsgn2 31504
Description: If 𝐴 <s 𝐵, then the sign of 𝐵 at the first place they differ is either undefined or 2𝑜. (Contributed by Scott Fenton, 4-Sep-2011.)
Assertion
Ref Expression
sltsgn2 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜)))
Distinct variable groups:   𝐴,𝑘   𝐵,𝑘

Proof of Theorem sltsgn2
StepHypRef Expression
1 sltval2 31498 . 2 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 ↔ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)})))
2 fvex 6160 . . . 4 (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
3 fvex 6160 . . . 4 (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
42, 3brtp 31338 . . 3 ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ↔ (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅) ∨ ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) ∨ ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜)))
5 orc 400 . . . . 5 ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
65adantl 482 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅) → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
7 olc 399 . . . . 5 ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜 → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
87adantl 482 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
97adantl 482 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
106, 8, 93jaoi 1388 . . 3 ((((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅) ∨ ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) ∨ ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜)) → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
114, 10sylbi 207 . 2 ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
121, 11syl6bi 243 1 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 383  wa 384  w3o 1035   = wceq 1480  wcel 1992  wne 2796  {crab 2916  c0 3896  {ctp 4157  cop 4159   cint 4445   class class class wbr 4618  Oncon0 5685  cfv 5850  1𝑜c1o 7499  2𝑜c2o 7500   No csur 31482   <s cslt 31483
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1841  ax-6 1890  ax-7 1937  ax-8 1994  ax-9 2001  ax-10 2021  ax-11 2036  ax-12 2049  ax-13 2250  ax-ext 2606  ax-sep 4746  ax-nul 4754  ax-pow 4808  ax-pr 4872  ax-un 6903
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3or 1037  df-3an 1038  df-tru 1483  df-ex 1702  df-nf 1707  df-sb 1883  df-eu 2478  df-mo 2479  df-clab 2613  df-cleq 2619  df-clel 2622  df-nfc 2756  df-ne 2797  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3193  df-sbc 3423  df-dif 3563  df-un 3565  df-in 3567  df-ss 3574  df-pss 3576  df-nul 3897  df-if 4064  df-pw 4137  df-sn 4154  df-pr 4156  df-tp 4158  df-op 4160  df-uni 4408  df-int 4446  df-br 4619  df-opab 4679  df-tr 4718  df-eprel 4990  df-po 5000  df-so 5001  df-fr 5038  df-we 5040  df-ord 5688  df-on 5689  df-suc 5691  df-iota 5813  df-fv 5858  df-1o 7506  df-2o 7507  df-slt 31486
This theorem is referenced by: (None)
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