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Proof step 6 from proof notone (previous step)

1 < i, i ^ j -1 * i < i, 0 < j, 2 ≤ i, i ^ j -1 * i ≤ 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 7:
i ^ j -1 < i, i ^ j -1 < j, i ^ j -1 < 1, 0 < j, 1 < i, i ^ j -1 + 1 < i, 2 ≤ i, i ^ j -1 ≤ 0, ∀ j0, i. i→n(j0) < i→n(j) → 1 < i ∧ 0 < j0 ∧ i ^ j0 ≤ 1 → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i +1 = i + 1; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%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% ⊦ 1 < i → (j * i < i ↔ j < 1); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0); (s)