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Proof step 15 from proof bytes2int-zero-06 (previous step)

∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1

User applied rule apply induction

And obtained the following new goals

Proof step 16:
∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ ∀ by. by ∈ bytes.butlast → b2i(by) = 0, bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1
Proof step 18:
bytes2int(bytes.butlast) = 0 ∨ bytes.butlast = [], ∀ by. by ∈ bytes → b2i(by) = 0, ∀ bytes0. # bytes0 < # bytes → (∀ by. by ∈ bytes0 → b2i(by) = 0) → bytes2int(bytes0) = 0 ∨ bytes0 = [] ⊦ bytes = [], bytes2int(bytes.butlast) * 256 + band(b2i(bytes.last), 255) = 0, # bytes = 1

Rule argument

subst: [bytes0] → [bytes.butlast]

Additional information

Used simplifier rules: ⊦ bytes ≠ [] → # bytes.butlast = # bytes - 1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bytes.butlast