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Proof step 6 from proof div-div-pred (previous step)

i0 = i1 + 2, i0 < 0, i0 < ~1 ⊦ (i0 / 2 - 1) / 2 = i1 / 4

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i1 < ~3, i1 < 0 ⊦ ((i1 + 2) / 2 - 1) / 2 = i1 / 4

Rule argument

Additional information

Used simplifier rules: numlsf: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward) from specification int-basic1%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* 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) < (* j0) ↔ i < j0 - j; (s)