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Theorem bj-peano4 10080
Description: Remove from peano4 4320 dependency on ax-setind 4262. Therefore, it only requires core constructive axioms (albeit more of them). (Contributed by BJ, 28-Nov-2019.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-peano4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))

Proof of Theorem bj-peano4
StepHypRef Expression
1 3simpa 901 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴 ∈ ω ∧ 𝐵 ∈ ω))
2 pm3.22 252 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐵 ∈ ω ∧ 𝐴 ∈ ω))
3 bj-nnen2lp 10079 . . . . 5 ((𝐵 ∈ ω ∧ 𝐴 ∈ ω) → ¬ (𝐵𝐴𝐴𝐵))
41, 2, 33syl 17 . . . 4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ¬ (𝐵𝐴𝐴𝐵))
5 sucidg 4153 . . . . . . . . . . . 12 (𝐵 ∈ ω → 𝐵 ∈ suc 𝐵)
6 eleq2 2101 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐵 ∈ suc 𝐴𝐵 ∈ suc 𝐵))
75, 6syl5ibrcom 146 . . . . . . . . . . 11 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵𝐵 ∈ suc 𝐴))
8 elsucg 4141 . . . . . . . . . . 11 (𝐵 ∈ ω → (𝐵 ∈ suc 𝐴 ↔ (𝐵𝐴𝐵 = 𝐴)))
97, 8sylibd 138 . . . . . . . . . 10 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐵𝐴𝐵 = 𝐴)))
109imp 115 . . . . . . . . 9 ((𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
11103adant1 922 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
12 sucidg 4153 . . . . . . . . . . . 12 (𝐴 ∈ ω → 𝐴 ∈ suc 𝐴)
13 eleq2 2101 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐴 ∈ suc 𝐴𝐴 ∈ suc 𝐵))
1412, 13syl5ibcom 144 . . . . . . . . . . 11 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵𝐴 ∈ suc 𝐵))
15 elsucg 4141 . . . . . . . . . . 11 (𝐴 ∈ ω → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
1614, 15sylibd 138 . . . . . . . . . 10 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐴𝐵𝐴 = 𝐵)))
1716imp 115 . . . . . . . . 9 ((𝐴 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
18173adant2 923 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
1911, 18jca 290 . . . . . . 7 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)))
20 eqcom 2042 . . . . . . . . 9 (𝐵 = 𝐴𝐴 = 𝐵)
2120orbi2i 679 . . . . . . . 8 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
2221anbi1i 431 . . . . . . 7 (((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2319, 22sylib 127 . . . . . 6 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
24 ordir 730 . . . . . 6 (((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2523, 24sylibr 137 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵))
2625ord 643 . . . 4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (¬ (𝐵𝐴𝐴𝐵) → 𝐴 = 𝐵))
274, 26mpd 13 . . 3 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → 𝐴 = 𝐵)
28273expia 1106 . 2 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
29 suceq 4139 . 2 (𝐴 = 𝐵 → suc 𝐴 = suc 𝐵)
3028, 29impbid1 130 1 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 97  wb 98  wo 629  w3a 885   = wceq 1243  wcel 1393  suc csuc 4102  ωcom 4313
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-13 1404  ax-14 1405  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022  ax-nul 3883  ax-pr 3944  ax-un 4170  ax-bd0 9933  ax-bdor 9936  ax-bdn 9937  ax-bdal 9938  ax-bdex 9939  ax-bdeq 9940  ax-bdel 9941  ax-bdsb 9942  ax-bdsep 10004  ax-infvn 10066
This theorem depends on definitions:  df-bi 110  df-3an 887  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-ral 2311  df-rex 2312  df-rab 2315  df-v 2559  df-dif 2920  df-un 2922  df-in 2924  df-ss 2931  df-nul 3225  df-sn 3381  df-pr 3382  df-uni 3581  df-int 3616  df-suc 4108  df-iom 4314  df-bdc 9961  df-bj-ind 10051
This theorem is referenced by: (None)
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