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

0 < i0, (i + i0 % j) + 1 < i0, i + 1 < i0, j ≤ i0, i + i0 % j < j, (i + i0 % j) + 1 < j, 0 ≤ i + i0, i < j, i < 0, 0 < j, i + 1 < j, 0 ≤ i0, i + i0 % j < 0 ⊦ 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-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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs