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Proof step 28 from proof fillsgn-butlast (previous step)

bytes = bytes0 + by ', 0 ≤ bytes2int(bytes0), bytes2int(bytes) < 0, # bytes ≤ n ⊦ fillsgn(n, bytes0) = mklist(i2b(~1), (n - # bytes) + 1) + bytes0, bytes = [], # bytes = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
bytes2int(bytes0 + by ') < 0, 0 ≤ bytes2int(bytes0), # bytes0 < n ⊦ bytes0 = [], fillsgn(n, bytes0) = mklist(i2b(~1), 1 + (n - # bytes0 + 1)) + bytes0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ m < n → n ≠ 0;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ bytes + by ' ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;