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Proof step 10 from proof div-div-pred-01 (previous step)

i - 1 = i0 * 2 + j0, i < 0, abs(j0) < 2, 0 < 2, i - 1 < 0, j0 ≤ 0, i0 ≤ 0 ⊦ (i0 - 3) / 4 = (i - 7) / 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
abs(j0) < 2, i0 * 2 + j0 < ~1, i0 * 2 + j0 < 0, j0 ≤ 0, i0 ≤ 0 ⊦ (i0 - 3) / 4 = (j0 + i0 * 2 - 6) / 8

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i + (j - i0) = i + j - i0; (s)%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ (* i) + i0 - (* j) = (i - j) + i0; (s)%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ i < 1 ↔ i ≤ 0; (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 < i0 ↔ i < j + i0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)