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Theorem bj-peano4 16550
Description: Remove from peano4 4695 dependency on ax-setind 4635. 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 1020 . . . . 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 16549 . . . . 5 ((𝐵 ∈ ω ∧ 𝐴 ∈ ω) → ¬ (𝐵𝐴𝐴𝐵))
41, 2, 33syl 17 . . . 4 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ¬ (𝐵𝐴𝐴𝐵))
5 sucidg 4513 . . . . . . . . . . . 12 (𝐵 ∈ ω → 𝐵 ∈ suc 𝐵)
6 eleq2 2295 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐵 ∈ suc 𝐴𝐵 ∈ suc 𝐵))
75, 6syl5ibrcom 157 . . . . . . . . . . 11 (𝐵 ∈ ω → (suc 𝐴 = suc 𝐵𝐵 ∈ suc 𝐴))
8 elsucg 4501 . . . . . . . . . . 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 1041 . . . . . . . 8 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → (𝐵𝐴𝐵 = 𝐴))
12 sucidg 4513 . . . . . . . . . . . 12 (𝐴 ∈ ω → 𝐴 ∈ suc 𝐴)
13 eleq2 2295 . . . . . . . . . . . 12 (suc 𝐴 = suc 𝐵 → (𝐴 ∈ suc 𝐴𝐴 ∈ suc 𝐵))
1412, 13syl5ibcom 155 . . . . . . . . . . 11 (𝐴 ∈ ω → (suc 𝐴 = suc 𝐵𝐴 ∈ suc 𝐵))
15 elsucg 4501 . . . . . . . . . . 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 1042 . . . . . . . 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 2233 . . . . . . . . 9 (𝐵 = 𝐴𝐴 = 𝐵)
2120orbi2i 769 . . . . . . . 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 824 . . . . . 6 (((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵) ↔ ((𝐵𝐴𝐴 = 𝐵) ∧ (𝐴𝐵𝐴 = 𝐵)))
2523, 24sylibr 134 . . . . 5 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ suc 𝐴 = suc 𝐵) → ((𝐵𝐴𝐴𝐵) ∨ 𝐴 = 𝐵))
2625ord 731 . . . 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 1231 . 2 ((𝐴 ∈ ω ∧ 𝐵 ∈ ω) → (suc 𝐴 = suc 𝐵𝐴 = 𝐵))
29 suceq 4499 . 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 715  w3a 1004   = wceq 1397  wcel 2202  suc csuc 4462  ωcom 4688
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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-13 2204  ax-14 2205  ax-ext 2213  ax-nul 4215  ax-pr 4299  ax-un 4530  ax-bd0 16408  ax-bdor 16411  ax-bdn 16412  ax-bdal 16413  ax-bdex 16414  ax-bdeq 16415  ax-bdel 16416  ax-bdsb 16417  ax-bdsep 16479  ax-infvn 16536
This theorem depends on definitions:  df-bi 117  df-3an 1006  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-rab 2519  df-v 2804  df-dif 3202  df-un 3204  df-in 3206  df-ss 3213  df-nul 3495  df-sn 3675  df-pr 3676  df-uni 3894  df-int 3929  df-suc 4468  df-iom 4689  df-bdc 16436  df-bj-ind 16522
This theorem is referenced by: (None)
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