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Proof step 2 from proof notone (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
2 ≤ i, i ^ j < i, i ^ j ≤ 1, 0 < j, 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Rule argument

Additional information

Used simplifier rules: numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%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