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Proof step 30 from proof bytes2bytes-fix (previous step)

0 ≤ b2i(int2bytes(bytes2int(by0 ' + bytes)).first), 0 ≤ b2i(by0), # int2bytes(bytes2int(by0 ' + bytes)) < 2 + # bytes, ∀ bytes0. # bytes0 < #(i2b(0) + by0 + bytes) → true → bytes0 = [] ∨ fix(# bytes0, int2bytes(bytes2int(bytes0))) = bytes0 ⊦ fix(1 + # bytes, int2bytes(bytes2int(by0 ' + bytes))) = by0 ' + bytes

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bytes0] → [by0 ' + bytes]

Additional information

Used simplifier rules: ⊦ by ' ≠ [];% ⊦ ¬ bytes + bytes0 ≠ [] ↔ bytes = [] ∧ bytes0 = [];% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ # by ' = 0 +1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by + bytes = by ' + bytes;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: by0 ' + bytes