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Proof step 16 from proof exdivmod (previous step)

i + j1 = j * i1, j1 + j < 0, 0 ≤ j1, 0 ≤ i1, i ≤ 0, j < 0 ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: i1, ~ j1 abort: F discard: T computed: F

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i;%neg-15: ⊦ 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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs Used terms: i1, ~ j1