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Proof step 7 from proof nummult-le-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
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

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%numadd: ⊦ i + (* j) = i0 + (* j0) ↔ i = i0 + (j0 - j); from specification int-abs