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Theorem bj-peano4 10446
Description: Remove from peano4 4347 dependency on ax-setind 4289. 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 912 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴 ∈ ω ∧ 𝐵 ∈ ω))
2 pm3.22 256 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐵 ∈ ω ∧ 𝐴 ∈ ω))
3 bj-nnen2lp 10445 . . . . 5 ((𝐵 ∈ ω ∧ 𝐴 ∈ ω) → ¬ (𝐵𝐴𝐴𝐵))
41, 2, 33syl 17 . . . 4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ¬ (𝐵𝐴𝐴𝐵))
5 sucidg 4180 . . . . . . . . . . . 12 (𝐵 ∈ ω → 𝐵 ∈ suc 𝐵)
6 eleq2 2117 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐵 ∈ suc 𝐴𝐵 ∈ suc 𝐵))
75, 6syl5ibrcom 150 . . . . . . . . . . 11 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵𝐵 ∈ suc 𝐴))
8 elsucg 4168 . . . . . . . . . . 11 (𝐵 ∈ ω → (𝐵 ∈ suc 𝐴 ↔ (𝐵𝐴𝐵 = 𝐴)))
97, 8sylibd 142 . . . . . . . . . 10 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐵𝐴𝐵 = 𝐴)))
109imp 119 . . . . . . . . 9 ((𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
11103adant1 933 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
12 sucidg 4180 . . . . . . . . . . . 12 (𝐴 ∈ ω → 𝐴 ∈ suc 𝐴)
13 eleq2 2117 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐴 ∈ suc 𝐴𝐴 ∈ suc 𝐵))
1412, 13syl5ibcom 148 . . . . . . . . . . 11 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵𝐴 ∈ suc 𝐵))
15 elsucg 4168 . . . . . . . . . . 11 (𝐴 ∈ ω → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
1614, 15sylibd 142 . . . . . . . . . 10 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐴𝐵𝐴 = 𝐵)))
1716imp 119 . . . . . . . . 9 ((𝐴 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
18173adant2 934 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
1911, 18jca 294 . . . . . . 7 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)))
20 eqcom 2058 . . . . . . . . 9 (𝐵 = 𝐴𝐴 = 𝐵)
2120orbi2i 689 . . . . . . . 8 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
2221anbi1i 439 . . . . . . 7 (((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2319, 22sylib 131 . . . . . 6 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
24 ordir 741 . . . . . 6 (((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2523, 24sylibr 141 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵))
2625ord 653 . . . 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 1117 . 2 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
29 suceq 4166 . 2 (𝐴 = 𝐵 → suc 𝐴 = suc 𝐵)
3028, 29impbid1 134 1 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wb 102  wo 639  w3a 896   = wceq 1259  wcel 1409  suc csuc 4129  ωcom 4340
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-bndl 1415  ax-4 1416  ax-13 1420  ax-14 1421  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444  ax-ext 2038  ax-nul 3910  ax-pr 3971  ax-un 4197  ax-bd0 10299  ax-bdor 10302  ax-bdn 10303  ax-bdal 10304  ax-bdex 10305  ax-bdeq 10306  ax-bdel 10307  ax-bdsb 10308  ax-bdsep 10370  ax-infvn 10432
This theorem depends on definitions:  df-bi 114  df-3an 898  df-tru 1262  df-nf 1366  df-sb 1662  df-clab 2043  df-cleq 2049  df-clel 2052  df-nfc 2183  df-ral 2328  df-rex 2329  df-rab 2332  df-v 2576  df-dif 2947  df-un 2949  df-in 2951  df-ss 2958  df-nul 3252  df-sn 3408  df-pr 3409  df-uni 3608  df-int 3643  df-suc 4135  df-iom 4341  df-bdc 10327  df-bj-ind 10417
This theorem is referenced by: (None)
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