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Theorem sltsgn2 31059
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 31053 . 2 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 ↔ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)})))
2 fvex 6113 . . . 4 (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
3 fvex 6113 . . . 4 (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
42, 3brtp 30892 . . 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 399 . . . . 5 ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
65adantl 481 . . . 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 398 . . . . 5 ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜 → ((𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜))
87adantl 481 . . . 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 481 . . . 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 1383 . . 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 206 . 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 242 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 382  wa 383  w3o 1030   = wceq 1475  wcel 1977  wne 2780  {crab 2900  c0 3874  {ctp 4129  cop 4131   cint 4410   class class class wbr 4583  Oncon0 5640  cfv 5804  1𝑜c1o 7440  2𝑜c2o 7441   No csur 31037   <s cslt 31038
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-8 1979  ax-9 1986  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590  ax-sep 4709  ax-nul 4717  ax-pow 4769  ax-pr 4833  ax-un 6847
This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3or 1032  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-eu 2462  df-mo 2463  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ne 2782  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-sbc 3403  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-pss 3556  df-nul 3875  df-if 4037  df-pw 4110  df-sn 4126  df-pr 4128  df-tp 4130  df-op 4132  df-uni 4373  df-int 4411  df-br 4584  df-opab 4644  df-tr 4681  df-eprel 4949  df-po 4959  df-so 4960  df-fr 4997  df-we 4999  df-ord 5643  df-on 5644  df-suc 5646  df-iota 5768  df-fv 5812  df-1o 7447  df-2o 7448  df-slt 31041
This theorem is referenced by: (None)
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