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Proof step 3 from proof div-div-pred (previous step)

i = i0 + 1, i < 0 ⊦ (i0 / 2 - 1) / 2 = (i - 3) / 4

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
i0 < ~1, i0 < 0 ⊦ (i0 / 2 - 1) / 2 = (i0 - 2) / 4

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; 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%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs