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Proof step 7 from proof square-nls (previous step)

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 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 8:
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) < j * (i0 + 1), j = 0

Rule argument

fma: j * (j + 1) < j * (i0 + 1)

Additional information

Used simplifier rules: ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult%int-simp-06: ⊦ i < i1 ∧ i2 ≤ i3 → (i1 + i3 < i + i2 ↔ false); from specification int-abs%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) < j * (i0 + 1)