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Proof step 16 from proof fillsgn-butlast (previous step)

bytes2int(bytes.butlast) < 0, bytes = [] ∨ 0 ≤ b2i(bytes.first), # bytes ≤ n ⊦ fillsgn(n, bytes.butlast) = mklist(i2b(0), (n - # bytes) + 1) + bytes.butlast, bytes = [], # bytes = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
0 ≤ b2i(bytes.first), # bytes ≤ n, bytes2int(bytes.butlast) < 0 ⊦ # bytes = 1, bytes = [], fillsgn(n, bytes.butlast) = mklist(i2b(0), (n - # bytes) + 1) + bytes.butlast

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)