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Proof step 14 from proof auxexdivmod (previous step)

i < (i0 + i2 -1) * j, 0 < j, i0 * j ≤ i, 0 ≤ i0, 0 ≤ i, 0 ≤ i2 -1, ∀ 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 all left

And obtained the following new goals

Proof step 15:
0 < j, i < (i0 + i2 -1) * j, 0 ≤ i2 -1, 0 ≤ i, 0 ≤ i0, i0 * j ≤ i, 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), ∀ 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

Rule argument

pos: left 7 terms: j, i, i0 abort: F discard: F computed: T

Additional information

Used terms: j, i, i0