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Proof step 15 from proof div-ls-mone (previous step)

~1 = i0 + i, i0 < i, i0 + 1 < i0 * j + 2 * j, 0 < i0 * j + 2 * j, i0 < 0, 1 + i0 * j < j, i0 < j, 0 < j, i0 + 1 < j, i0 < i0 * j + 2 * j, 0 < i, i0 + 1 < i ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: ~ i + ~1 = i0

Additional information

Used simplifier rules: neg-15: ⊦ i + ~ j = i - j; 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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs Used formulas: ~ i + ~1 = i0