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Proof step 4 from proof int2bits-mod256 (previous step)

0 < j, 0 ≤ j0, j0 < 2 ^ 8 ⊦ int2bits(j * 2 ^ 8 + j0) = int2bits(j) + fillsgn(9, int2bits(j0)).rest

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < j ∧ 0 ≤ j0 ∧ j0 < 2 ^ i ∧ 0 ≤ i → int2bits(j * 2 ^ i + j0) = int2bits(j) + fillsgn(i→n(i + 1), int2bits(j0)).rest

Rule argument

name: int2bits-mod2pot rotate: F subst: [j0, i, j] → [j0, 8, j]

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0); Used terms: j0, 8, j