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Proof step 6 from proof ls-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i0 ^ j < i ^ j, 0 ≤ i0, 0 ≤ i, 0 ≤ j, i ≤ i0, ∀ j0, i0, i. i→n(j0) < i→n(j) → i0 ^ j0 < i ^ j0 ∧ i ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ i ∧ 0 ≤ i0 → false ⊦ i0 ^ (j - 1) < i ^ (j - 1) ∧ 0 < j

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 1 ≤ i ↔ 0 < i; (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)