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Proof step 3 from proof notone (previous step)

1 < i, 0 < j, i ^ j ≤ 1, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i ^ j < i, 0 < j, 1 < i, i ^ j ≤ 1, 2 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

Additional information

Used simplifier rules: numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs% ⊦ 1 ≤ i ↔ 0 < i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs