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Proof step 1 from proof bytes2int-len-one-02 (conclusion)

⊦ 0 ≤ bytes2int(bytes) ∧ # bytes ≠ 1 ∧ bytes ≠ [] ∧ bytes.first = i2b(0) ∧ b2i(bytes.rest.first) < 0 → ¬ bytes2int(bytes) < 128

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
bytes.first = i2b(0), b2i(bytes.rest.first) < 0, 0 ≤ bytes2int(bytes), bytes2int(bytes) < 128 ⊦ # bytes = 1, bytes = []

Rule argument

Additional information

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