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Proof step 16 from proof bytes2int-zero-03 (previous step)

ub2i(by0) = 0, b2i(by0) = 0, bytes2int(bytes1) = 0, by ∈ bytes1, ∀ by, bytes0. # bytes0 < #(bytes1 + by0 ') → bytes2int(bytes0) = 0 ∧ by ∈ bytes0 → by = i2b(0) ⊦ by = by0, bytes1 = [], by = i2b(0)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [by, bytes0] → [by, bytes1]

Additional information

Used simplifier rules: ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: by, bytes1