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Proof step 10 from proof int2bits-mod2-neg (previous step)

abs(j) < 2, j ≤ 0, i0 ≤ 0, i0 * 2 + j < 0 ⊦ i0 * 2 + j = ~1, int2bits(j) = (~1 = j)'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ j < 0 → abs(j) = 0 - j

Rule argument

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

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';%int2bits-true: ⊦ int2bits(j) = true ' ↔ j = ~1;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;%j0 ≠ 0, j < j0 ⊦ i ≤ (* j) → i < (* j0);% ⊦ ¬ j < 0 → abs(j) = j;% ⊦ 0 - i = ~ i;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ i < 0 → (~2 < i ↔ i = ~1);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: j