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Proof step 6 from proof fillsgn-empty (previous step)

¬(# bytes < n ∧ 0 ≤ bytes2int(bytes)), fillsgn(n, bytes) = [], # bytes < n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
fillsgn(n, bytes) = [], # bytes < n, bytes2int(bytes) < 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;