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Proof step 12 from proof square-nls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: numlelsdel: j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); from specification int-abs%two-ls: ⊦ 2 * i + j < 2 * (i + i0) ↔ j < 2 * i0; from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%ls-forward: ⊦ i * j < i * j0 → i ≠ 0; (forward) from specification int-mult%lels-01: ⊦ ¬ i ≤ j ↔ j < i; 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%int-simp-06: ⊦ i < i1 ∧ i2 ≤ i3 → (i1 + i3 < i + i2 ↔ false); from specification int-abs%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs