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Proof step 5 from proof notone (previous step)

0 < j, 2 ≤ i, i ^ j -1 * i < i, i ^ j -1 * i ≤ 1, 0 < j, 1 < i, ∀ j0, i. i→n(j0) < i→n(j) → true → 1 < i ∧ 0 < j0 → ¬ i ^ j0 ≤ 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i ^ j -1 < i, i ^ j -1 ≤ 0, i ^ j -1 < 1, 2 ≤ i, 1 < i, 0 < j, i ^ j -1 < j, i ^ j -1 + 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: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%le-one: ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%ls-same-05: ⊦ 1 < i → (j * i < i ↔ j < 1); from specification int-mult