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Proof step 4 from proof div-mod (previous step)

0 < j, 0 ≤ i0, 0 ≤ i ⊦ j | i - i0 ↔ i % j = i0 % j, 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

User applied rule cut formula

And obtained the following new goals

Proof step 5:
i - i0 = ~(i0 - i), 0 < j, 0 ≤ i, 0 ≤ i0 ⊦ j | i - i0 ↔ i % j = i0 % j, 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

Rule argument

fma: i - i0 = ~(i0 - i)

Additional information

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