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Proof step 6 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 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 7:
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

Rule argument

fma: j = 0

Additional information

Used simplifier rules: 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%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used formulas: j = 0