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Theorem sucneqond 38493
Description: Inequality of an ordinal set with its successor. Does not use the axiom of regularity. (Contributed by ML, 18-Oct-2020.)
Hypotheses
Ref Expression
sucneqond.1 (𝜑 → 𝑋 = suc 𝑌)
sucneqond.2 (𝜑 → 𝑌 ∈ On)
Assertion
Ref Expression
sucneqond (𝜑 → 𝑋 ≠ 𝑌)

Proof of Theorem sucneqond
StepHypRef Expression
1 sucneqond.2 . . . . 5 (𝜑 → 𝑌 ∈ On)
2 sucidg 6467 . . . . 5 (𝑌 ∈ On → 𝑌 ∈ suc 𝑌)
31, 2syl 18 . . . 4 (𝜑 → 𝑌 ∈ suc 𝑌)
4 sucneqond.1 . . . 4 (𝜑 → 𝑋 = suc 𝑌)
53, 4eleqtrrd 2867 . . 3 (𝜑 → 𝑌 ∈ 𝑋)
6 onsuc 7851 . . . . . . . 8 (𝑌 ∈ On → suc 𝑌 ∈ On)
71, 6syl 18 . . . . . . 7 (𝜑 → suc 𝑌 ∈ On)
84, 7eqeltrd 2864 . . . . . 6 (𝜑 → 𝑋 ∈ On)
9 eloni 6393 . . . . . 6 (𝑋 ∈ On → Ord 𝑋)
108, 9syl 18 . . . . 5 (𝜑 → Ord 𝑋)
11 ordirr 6401 . . . . 5 (Ord 𝑋 → ¬ 𝑋 ∈ 𝑋)
1210, 11syl 18 . . . 4 (𝜑 → ¬ 𝑋 ∈ 𝑋)
13 eleq1 2852 . . . . . 6 (𝑋 = 𝑌 → (𝑋 ∈ 𝑋 ↔ 𝑌 ∈ 𝑋))
1413biimprd 251 . . . . 5 (𝑋 = 𝑌 → (𝑌 ∈ 𝑋 → 𝑋 ∈ 𝑋))
1514con3d 153 . . . 4 (𝑋 = 𝑌 → (¬ 𝑋 ∈ 𝑋 → ¬ 𝑌 ∈ 𝑋))
1612, 15syl5com 32 . . 3 (𝜑 → (𝑋 = 𝑌 → ¬ 𝑌 ∈ 𝑋))
175, 16mt2d 137 . 2 (𝜑 → ¬ 𝑋 = 𝑌)
1817neqned 2966 1 (𝜑 → 𝑋 ≠ 𝑌)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   = wceq 1570   ∈ wcel 2147   ≠ wne 2959  Ord word 6382  Oncon0 6383  suc csuc 6385
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735  ax-sep 5253  ax-pr 5396  ax-un 7778
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3or 1104  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-pss 4017  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4839  df-uni 4871  df-br 5107  df-opab 5172  df-tr 5217  df-eprel 5556  df-po 5564  df-so 5565  df-fr 5609  df-we 5611  df-ord 6386  df-on 6387  df-suc 6389
This theorem is used by:  sucneqoni  38494
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