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Theorem bj-peano4 16725
Description: Remove from peano4 4719 dependency on ax-setind 4659. 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 1021 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴 ∈ ω ∧ 𝐵 ∈ ω))
2 pm3.22 265 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (𝐵 ∈ ω ∧ 𝐴 ∈ ω))
3 bj-nnen2lp 16724 . . . . 5 ((𝐵 ∈ ω ∧ 𝐴 ∈ ω) → ¬ (𝐵𝐴𝐴𝐵))
41, 2, 33syl 17 . . . 4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ¬ (𝐵𝐴𝐴𝐵))
5 sucidg 4537 . . . . . . . . . . . 12 (𝐵 ∈ ω → 𝐵 ∈ suc 𝐵)
6 eleq2 2296 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐵 ∈ suc 𝐴𝐵 ∈ suc 𝐵))
75, 6syl5ibrcom 157 . . . . . . . . . . 11 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵𝐵 ∈ suc 𝐴))
8 elsucg 4525 . . . . . . . . . . 11 (𝐵 ∈ ω → (𝐵 ∈ suc 𝐴 ↔ (𝐵𝐴𝐵 = 𝐴)))
97, 8sylibd 149 . . . . . . . . . 10 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐵𝐴𝐵 = 𝐴)))
109imp 124 . . . . . . . . 9 ((𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
11103adant1 1042 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
12 sucidg 4537 . . . . . . . . . . . 12 (𝐴 ∈ ω → 𝐴 ∈ suc 𝐴)
13 eleq2 2296 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐴 ∈ suc 𝐴𝐴 ∈ suc 𝐵))
1412, 13syl5ibcom 155 . . . . . . . . . . 11 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵𝐴 ∈ suc 𝐵))
15 elsucg 4525 . . . . . . . . . . 11 (𝐴 ∈ ω → (𝐴 ∈ suc 𝐵 ↔ (𝐴𝐵𝐴 = 𝐵)))
1614, 15sylibd 149 . . . . . . . . . 10 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵 → (𝐴𝐵𝐴 = 𝐵)))
1716imp 124 . . . . . . . . 9 ((𝐴 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
18173adant2 1043 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐴𝐵𝐴 = 𝐵))
1911, 18jca 306 . . . . . . 7 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)))
20 eqcom 2234 . . . . . . . . 9 (𝐵 = 𝐴𝐴 = 𝐵)
2120orbi2i 770 . . . . . . . 8 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
2221anbi1i 458 . . . . . . 7 (((𝐵𝐴𝐵 = 𝐴) ∧ (𝐴𝐵𝐴 = 𝐵)) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2319, 22sylib 122 . . . . . 6 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
24 ordir 825 . . . . . 6 (((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2523, 24sylibr 134 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵))
2625ord 732 . . . 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 1232 . 2 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
29 suceq 4523 . 2 (𝐴 = 𝐵 → suc 𝐴 = suc 𝐵)
3028, 29impbid1 142 1 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 104  wb 105  wo 716  w3a 1005   = wceq 1398  wcel 2203  suc csuc 4486  ωcom 4712
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-13 2205  ax-14 2206  ax-ext 2214  ax-nul 4236  ax-pr 4322  ax-un 4554  ax-bd0 16583  ax-bdor 16586  ax-bdn 16587  ax-bdal 16588  ax-bdex 16589  ax-bdeq 16590  ax-bdel 16591  ax-bdsb 16592  ax-bdsep 16654  ax-infvn 16711
This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-ral 2525  df-rex 2526  df-rab 2529  df-v 2815  df-dif 3213  df-un 3215  df-in 3217  df-ss 3224  df-nul 3509  df-sn 3695  df-pr 3696  df-uni 3915  df-int 3950  df-suc 4492  df-iom 4713  df-bdc 16611  df-bj-ind 16697
This theorem is referenced by: (None)
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