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Proof step 2 from proof square-ls-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: (j + 1) * (j + 2) = (j + 1) * ((j + 1) + 1)

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%numsucc: ⊦ (* i0) * i + i = (i0 + 1) * i;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s) Used formulas: (j + 1) * (j + 2) = (j + 1) * ((j + 1) + 1)