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Proof step 10 from proof bytes2int-zero-06 (previous step)

¬ 1 < # bytes, ∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# bytes = 1, ∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < 1 → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes) = 0

Rule argument

Additional information

Used simplifier rules: ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);