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Proof step 4 from proof int2bytes-len-le-neg-03 (previous step)

0 ≤ b2i(int2bytes(i).rest.first) → ~(2 ^ ((n→i(# int2bytes(i).rest.rest) + 1) * 8)) ≤ bytes2int(i2b(~1) ' + int2bytes(i).rest.first ' + int2bytes(i).rest.rest), int2bytes(i).first = i2b(~1) ⊦ # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n) - 1) + i < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
int2bytes(i).first = i2b(~1), ¬ b2i(int2bytes(i).rest.first) < 0 → 0 ≤ 2 ^ ((n→i(# int2bytes(i).rest.rest) + 1) * 8) + bytes2int(i2b(~1) ' + int2bytes(i).rest.first ' + int2bytes(i).rest.rest) ⊦ # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n) - 1) + i < 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i ≤ j ↔ 0 ≤ i + j;% ⊦ ¬ i ≤ j ↔ j < i;