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Proof step 2 from proof div-add-nosucc-neg (previous step)

i % j = 0, i < j, 0 < j, 1 < j, i < 0, j + i ≤ 0, i + 1 < j ⊦ j = 1, (i + 1 - j) / j = i / j

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i + 1 - j = ~(~ i + (j - 1)), i % j = 0, i < j, 0 < j, 1 < j, i < 0, j + i ≤ 0, i + 1 < j ⊦ j = 1, (i + 1 - j) / j = i / j

Rule argument

fma: i + 1 - j = ~(~ i + (j - 1))

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%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-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%minus-sub: ⊦ ~(i - j) = j - i; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2 Used formulas: i + 1 - j = ~(~ i + (j - 1))