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Proof step 20 from proof modmod-add-if (previous step)

i + i0 % j < 0, i < 0, 0 < j, 0 < i0, 0 ≤ i0, 0 ≤ i + i0, j ≤ i0 ⊦ (i + i0 % j) % j = ((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0), j = 1

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: modmod-add2-negpos-pos-01: ⊦ i + i0 % j < 0 ∧ 0 ≤ i + i0 ∧ i < 0 ∧ 0 ≤ i0 ∧ (i + i0 % j) % j ≠ 0 ∧ 0 < j → (i + i0 % j) % j = (i + i0) % j - j;%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)