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Proof step 19 from proof div-shift-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-zero: ⊦ i + 0 = i; from specification int-basic2%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ge-add-01: ⊦ 0 ≤ j ∧ 0 ≤ j0 ∧ 0 ≤ i0 → 0 ≤ i0 * j + j0; from specification int-mult%lstwo: ⊦ i ≠ 1 ∧ 0 ≤ i → (i < 2 ↔ i = 0); from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult