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Proof step 4 from proof int2bytes-fillsgn-rec (previous step)

¬(i < ~128 ∨ 128 ≤ i), # int2bytes(i) ≤ n ⊦ fillsgn(n, int2bytes(i)) = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) ', # int2bytes(i) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# int2bytes(i) ≤ n, ~128 ≤ i, i < 128 ⊦ fillsgn(n, int2bytes(i)) = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) ', # int2bytes(i) = 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;