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Proof step 2 from proof mod-sub (previous step)

0 < i, j < 0 ⊦ (i - j) % i = ~(j % i)

User applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: i - j = i + ~ j

Additional information

Used simplifier rules: ⊦ 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 = i0 ↔ i = i0 + j; (s)% ⊦ i + (j - i) = j; (s)% ⊦ ~ i = j ↔ i + j = 0; (s) Used formulas: i - j = i + ~ j