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Proof step 7 from proof ls-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i0 ^ (i1 + 1) < i ^ (i1 + 1), i ≤ i0, 0 ≤ i0, 0 ≤ i, ~1 ≤ i1, ∀ j0, i0, i. i→n(j0) < i→n(i1 + 1) → i0 ^ j0 < i ^ j0 ∧ i ≤ i0 ∧ 0 ≤ i0 ∧ 0 ≤ i ∧ 0 ≤ j0 → false ⊦ i0 ^ i1 < i ^ i1 ∧ 0 ≤ i1

Rule argument

Additional information

Used simplifier rules: mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-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%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs