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Proof step 16 from proof int2bytes-rec-pos-04 (previous step)

¬(# i2b(i % 256) ' < 2 ∧ 0 ≤ bytes2int(i2b(i % 256) ')), i % 256 ≤ 127, 256 ≤ i, 0 < i, 0 ≤ i ⊦ i2b(i % 256) ' = fix(2, i2b(i % 256) ').rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
b2i(i2b(i % 256)) < 0, 0 ≤ i, 0 < i, 256 ≤ i, i % 256 ≤ 127 ⊦ i2b(i % 256) ' = fix(2, i2b(i % 256) ').rest

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ # by ' = 0 +1;