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Proof step 5 from proof mod-two-pred-01 (previous step)

2 * i = (i0 * 2 + j0) + 1, abs(j0) < 2, i < 0, j0 ≤ 0, i0 ≤ 0 ⊦ j0 = ~1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j0] flags: rewrite

Additional information

Used simplifier rules: neg-06: ⊦ 0 - i = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%minus-one: ⊦ i < 0 → (~2 < i ↔ i = ~1); from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%numlsledel: j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%odd-even: ⊦ 2 * i ≠ 2 * j + 1; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j0