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Proof step 21 from proof auxexdivmod (previous step)

0 < j, i < (i0 + i2 -1) * j, i < (i1 + 1) * j, 0 ≤ i2 -1, 0 ≤ i, 0 ≤ i0, i0 * j ≤ i, i1 * j ≤ i, 0 ≤ i1, ∀ j, i, i0. 0 < j ∧ i < (i0 + i2) * j ∧ 0 ≤ i2 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 * j ≤ i → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0) ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: i1 abort: F discard: T computed: T

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;% ⊦ i -1 = i - 1; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i + (j - i0) = i + j - i0; (s) Used terms: i1