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Proof step 6 from proof fix-length (previous step)

¬(# bytes < n ∧ bytes2int(bytes) < 0), # bytes < n ⊦ # fix(n, bytes) = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bytes < n, 0 ≤ bytes2int(bytes) ⊦ # fix(n, bytes) = n

Rule argument

Additional information

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