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Proof step 2 from proof minus-add (previous step)

⊦ (i - j) / j0 + (j - i) / j0 = 0, j0 = 0

User applied rule cut formula

And obtained the following new goals

Proof step 3:
j - i = ~(i - j) ⊦ (i - j) / j0 + (j - i) / j0 = 0, j0 = 0

Rule argument

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

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; 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-def: ⊦ i + (j - i) = j; from specification int-abs Used formulas: j - i = ~(i - j)