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Proof step 5 from proof int2bits-add-high (previous step)

j < 0, j0 < 0, 0 < i, 2 ^ (i - 1) + j0 < 0, 0 < 2 ^ i + j0, 0 ≤ j * 2 ^ i + j0 ⊦ int2bits(j * 2 ^ i + j0) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0).rest)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < 0 ∧ j0 < j ∧ 0 < j → ¬ 0 ≤ i * j + j0

Rule argument

seq: ⊦ i < 0 ∧ j0 < j ∧ 0 < j → ¬ 0 ≤ i * j + j0 subst: [j0, j, i] → [j0, 2 ^ i, j] rotate: F spec: int-mult inst: name: modzero-03

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ j ≤ i ∧ i < 0 → j < 0; (forward)% ⊦ ¬ 2 ^ i < 0; Used terms: j0, 2 ^ i, j