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Proof step 14 from proof nummult-le-02 (previous step)

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

Heuristic var elimination applied rule insert elim lemma

And closed the goal!

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: eq-succ: ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1, 1, i2