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Proof step 10 from proof div2-add-neg (previous step)

~1 = i0 + j, j % 2 = 0, j < 0, i0 < 0 ⊦ ((i0 + 1) + j) / 2 = i0 / 2 + j / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: numadd-01 seq: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0 subst: [j0, i0, j] → [j, i0, ~1] flags: rewrite

Additional information

Used simplifier rules: Sub-def: ⊦ i + (j - i) = j; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-zero: ⊦ i + 0 = i; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: j, i0, ~1