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Proof step 6 from proof div-div-pred-01 (previous step)

(i - 1) / 2 < 0, i < 0, i < 0 ⊦ ((i - 1) / 2 - 3) / 4 = (i - 7) / 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i < 0 ⊦ ((i - 1) / 2 - 3) / 4 = (i - 7) / 8

Rule argument

Additional information

Used simplifier rules: div-ls-zero: 0 < j ⊦ i / (* j) < 0 ↔ i + j ≤ 0;%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%mone-05: ⊦ i ≤ ~1 ↔ i < 0; from specification int-abs%numle-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%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs