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Proof step 3 from proof bytes2int-len-one-02 (previous step)

bytes = by + bytes0, by = i2b(0), b2i(bytes0.first) < 0, 0 ≤ bytes2int(bytes), bytes2int(bytes) < 128 ⊦ # bytes = 1, bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 ≤ bytes2int(i2b(0) ' + bytes0), b2i(bytes0.first) < 0, bytes2int(i2b(0) ' + bytes0) < 128 ⊦ bytes0 = []

Rule argument

Additional information

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