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Proof step 1 from proof int2bits-256-neg-mone (conclusion)

⊦ 0 < j ∧ 0 < i → int2bits(j * 256 - 1) = int2bits(j - 1) + mklist(true, i→n(8))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 < j, 0 < i ⊦ int2bits(j * 256 - 1) = int2bits(j - 1) + true ' + true ' + true ' + true ' + true ' + true ' + true ' + true '

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(true, (* n)) = true ' + mklist(true, n -1);