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Proof step 20 from proof nummult-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i + (i1 - 1) * i = i1 * i, 0 < i1, i0 < i1, 0 < i, j0 < i, (i1 - 1) * i + j0 < i * i0, 0 ≤ i + (i1 - 1) * i + j0, 0 ≤ j0, 0 ≤ i0, ~1 ≤ (i1 - 1) * i + j0 ⊦

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i + 1 ≤ j ↔ i < j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)