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Proof step 2 from proof square-nls-01 (previous step)

i < 0, 0 ≤ j, (j * (j + 1)) / 2 < i + j ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i < 0, 0 ≤ j, (j * (j + 1)) / 2 < i + j ⊦ (j * (j + 1)) / 2 < j

Rule argument

fma: (j * (j + 1)) / 2 < j

Additional information

Used simplifier rules: square-nls: ⊦ 0 ≤ j → ¬ (j * (j + 1)) / 2 < j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used formulas: (j * (j + 1)) / 2 < j