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Proof step 23 from proof byte2bits-val (previous step)

b2i(by) < 0, 2 ^ n→i(n) = i + 1, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ (2 ^ n→i(# int2bits(b2i(by))) + b2i(by)) + 2 ^ n→i(# int2bits(b2i(by))) * i = b2i(by) + 256, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
2 ^ n→i(n) = i + 1, 8 = n + # int2bits(b2i(by)), # int2bits(b2i(by)) ≤ 8, b2i(by) < 0 ⊦ n = 0, 2 ^ n→i(# int2bits(b2i(by))) * (i + 1) = 256

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i = i * (j + 1);