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Proof step 5 from proof int2bits-rec-2pot-neg-01 (previous step)

0 < i, 2 ^ i * (j0 + 1) ≤ 0, 2 ^ i * (j0 + 1) < i ⊦ int2bits(j0 * 2 ^ i) = int2bits(j0) + mklist(false, i→n(i))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
0 ≤ i, j ≠ 0 ⊦ int2bits(j * 2 ^ i) = int2bits(j) + mklist(false, i→n(i))

Rule argument

name: int2bits-2pot-04 rotate: F subst: [i, j] → [i, j0]

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ i * 0 = 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ 1 * i = i;% ⊦ 2 ^ i ≤ 0 ↔ i < 0; Used terms: i, j0