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Proof step 5 from proof exdivmodpos (previous step)

0 ≤ i0, 0 < j, 0 ≤ i, i < (i0 + 1) * j, i0 * j ≤ i ⊦ ∃ i0, j0. i = i0 * j + j0 ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ j0 < j

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i0, i - i0 * j abort: F discard: F computed: F

Additional information

Used simplifier rules: dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs Used terms: i0, i - i0 * j