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Proof step 3 from proof div-mod (previous step)

¬(0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i) ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)

User applied rule cut formula

And obtained the following new goals

Proof step 4:
i - i0 = ~(i0 - i), ¬(0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i) ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)

Rule argument

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

Additional information

Used simplifier rules: 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%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)