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Proof step 20 from proof div-shift-two (previous step)

i0 = i1 * 2 + j1, 0 < j, j0 < j, j1 < 2, 0 < 2, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦ i0 = 1 + 2 * i1, j = 0, i0 * j + j0 < (1 + 2 * i1) * j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i + 0 = i; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≠ 1 ∧ 0 ≤ i → (i < 2 ↔ i = 0); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0; (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)%commutativity: ⊦ i * j = j * i; from specification int-mult