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Proof step 9 from proof le-pos-01 (previous step)

0 < i, i0 < i, 0 < j, i ^ j ≤ i0 ^ j, 0 ≤ j, 0 ≤ i0, 0 ≤ i, ∀ j0, i, i0. i→n(j0) < i→n(j) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦ 0 < i ∧ i0 < i ∧ 0 < j - 1 ∧ i ^ (j - 1) ≤ i0 ^ (j - 1) ∧ 0 ≤ j - 1 ∧ 0 ≤ i0 ∧ 0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
0 < j, i0 < i, 0 < i, 0 ≤ i, 0 ≤ i0, 0 ≤ j, i ^ j ≤ i0 ^ j, ∀ j0, i, i0. i→n(j0) < i→n(j) → 0 < i ∧ i0 < i ∧ 0 < j0 ∧ i ^ j0 ≤ i0 ^ j0 ∧ 0 ≤ j0 ∧ 0 ≤ i0 ∧ 0 ≤ i → false ⊦ 1 < j ∧ i ^ (j - 1) ≤ i0 ^ (j - 1)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < j - i0 ↔ i + i0 < j; (s)