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Proof step 3 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 4:
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: ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + ~ j = i - j; (s)%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) = j; (s) Used formulas: j - i = ~(i - j)