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Proof step 12 from proof auxexdivmod (previous step)

i1 * j ≤ i ∧ i < (i1 + 1) * j ∧ 0 ≤ i1, ∀ i0, j, i. i0 * j ≤ i ∧ i < (i0 + i2) * j ∧ 0 ≤ i0 ∧ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i2 → (∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0), i0 * j ≤ i, i < (i0 + i2 +1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i2 +1 ⊦ ∃ 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;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs Used terms: i1