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Proof step 4 from proof auxexdivmod-01 (previous step)

0 < j, i * j ≤ i, 0 ≤ i ⊦ ∃ i0. i0 * j ≤ i ∧ i < i0 * j + j ∧ 0 ≤ i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1)
Proof step 6:
0 < j, 0 ≤ i, i * j ≤ i ⊦ 0 < i, ∃ i0. i0 * j ≤ i ∧ i < i0 * j + j ∧ 0 ≤ i0
Proof step 8:
0 < j, 0 < i, 0 ≤ i, j ≤ 1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < i0 * j + j ∧ 0 ≤ i0

Rule argument

spec: <toplevel> name: le-same seq: ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1) subst: [j, i] → [j, i] flags: <all paths>

Additional information

Used terms: j, i