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Theorem sltsgn1 31512
Description: If 𝐴 <s 𝐵, then the sign of 𝐴 at the first place they differ is either undefined or 1𝑜. (Contributed by Scott Fenton, 4-Sep-2011.)
Assertion
Ref Expression
sltsgn1 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜)))
Distinct variable groups:   𝐴,𝑘   𝐵,𝑘

Proof of Theorem sltsgn1
StepHypRef Expression
1 sltval2 31507 . 2 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 ↔ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)})))
2 fvex 6158 . . . 4 (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
3 fvex 6158 . . . 4 (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) ∈ V
42, 3brtp 31344 . . 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 olc 399 . . . . 5 ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
65adantr 481 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅) → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
75adantr 481 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜 ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
8 orc 400 . . . . 5 ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
98adantr 481 . . . 4 (((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∧ (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 2𝑜) → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
106, 7, 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)) → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
114, 10sylbi 207 . 2 ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}){⟨1𝑜, ∅⟩, ⟨1𝑜, 2𝑜⟩, ⟨∅, 2𝑜⟩} (𝐵 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜))
121, 11syl6bi 243 1 ((𝐴 No 𝐵 No ) → (𝐴 <s 𝐵 → ((𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = ∅ ∨ (𝐴 {𝑘 ∈ On ∣ (𝐴𝑘) ≠ (𝐵𝑘)}) = 1𝑜)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wo 383  wa 384  w3o 1035   = wceq 1480  wcel 1987  wne 2790  {crab 2911  c0 3891  {ctp 4152  cop 4154   cint 4440   class class class wbr 4613  Oncon0 5682  cfv 5847  1𝑜c1o 7498  2𝑜c2o 7499   No csur 31491   <s cslt 31492
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 1836  ax-6 1885  ax-7 1932  ax-8 1989  ax-9 1996  ax-10 2016  ax-11 2031  ax-12 2044  ax-13 2245  ax-ext 2601  ax-sep 4741  ax-nul 4749  ax-pow 4803  ax-pr 4867  ax-un 6902
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 1878  df-eu 2473  df-mo 2474  df-clab 2608  df-cleq 2614  df-clel 2617  df-nfc 2750  df-ne 2791  df-ral 2912  df-rex 2913  df-rab 2916  df-v 3188  df-sbc 3418  df-dif 3558  df-un 3560  df-in 3562  df-ss 3569  df-pss 3571  df-nul 3892  df-if 4059  df-pw 4132  df-sn 4149  df-pr 4151  df-tp 4153  df-op 4155  df-uni 4403  df-int 4441  df-br 4614  df-opab 4674  df-tr 4713  df-eprel 4985  df-po 4995  df-so 4996  df-fr 5033  df-we 5035  df-ord 5685  df-on 5686  df-suc 5688  df-iota 5810  df-fv 5855  df-1o 7505  df-2o 7506  df-slt 31495
This theorem is referenced by: (None)
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