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Proof step 8 from proof nummult-le-02 (previous step)

i + (i1 - 1) * i = i1 * i, 0 < i, 0 ≤ i1, 0 ≤ j0, j0 < i, ~1 ≤ (i1 - 1) * i + j0, 0 ≤ i + (i1 - 1) * i + j0 ⊦ (i1 - 1) * i + j0 < i * i0 ↔ i1 ≤ i0

Heuristic var elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
i1 = i2 + 1, i + i2 * i = i1 * i, 0 < i, 0 ≤ i1, 0 ≤ j0, j0 < i, ~1 ≤ i2 * i + j0, 0 ≤ i + i2 * i + j0 ⊦ i2 * i + j0 < i * i0 ↔ i1 ≤ i0

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i1, 1, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1, 1, i2