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Proof step 4 from proof bytes2int-len-one-01 (previous step)

bytes = bytes0 + by ', bytes = bytes0 + by, bytes2int(bytes) < 128, 0 ≤ bytes2int(bytes) ⊦ bytes.first = i2b(0), bytes = [], # bytes = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
bytes2int(bytes0 + by ') < 128, 0 ≤ bytes2int(bytes0 + by ') ⊦ bytes0 = [], bytes0.first = i2b(0)

Rule argument

Additional information

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