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Proof step 6 from proof square-nle (previous step)

j * (j + 1) = i0 * 2 + j0 ∧ 0 ≤ j0 ∧ j0 < 2 ∧ 0 ≤ i0 ∧ 0 < 2 ∧ 0 ≤ j * (j + 1), i0 ≤ j, 0 ≤ j, 2 ≤ j, 0 < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j * (j + 1) = i0 * 2 + j0, j0 < 2, i0 ≤ j, 0 ≤ j, 2 ≤ j, 0 < j, 0 ≤ j0, j0 < j, 0 ≤ i0 ⊦

Rule argument

Additional information

Used simplifier rules: forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%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%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult