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Proof step 10 from proof modmod-add-if (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: modmod-add2-posneg-neg-01: ⊦ 0 ≤ i + i0 % j ∧ i + i0 < 0 ∧ 0 ≤ i ∧ i0 < 0 ∧ (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%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2