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Proof step 10 from proof le-pos-01 (previous step)

j = i1 + 1, 0 < j, 0 ≤ i0, 0 ≤ i, i ^ j ≤ i0 ^ j, i0 < i, 0 ≤ j, 0 < i, ∀ j0, i0, i. i→n(j0) < i→n(j) → 0 < j0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ i ^ j0 ≤ i0 ^ j0 ∧ i0 < i ∧ 0 ≤ j0 ∧ 0 < i → false ⊦ 1 < j ∧ i ^ i1 ≤ i0 ^ i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 < i, 0 < i, 0 ≤ i0, 0 ≤ i, i ^ i1 * i ≤ i0 ^ i1 * i0, 0 ≤ i1, ∀ j0, i0, i. i→n(j0) < i→n(i1 + 1) → 0 < j0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ i ^ j0 ≤ i0 ^ j0 ∧ i0 < i ∧ 0 ≤ j0 ∧ 0 < i → false ⊦ 0 < i1 ∧ i ^ i1 ≤ i0 ^ i1

Rule argument

Additional information

Used simplifier rules: succ: ⊦ 0 ≤ i → j ^ (i + 1) = j ^ i * j;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs