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Proof step 20 from proof bytes2int-mklist-mones-01 (previous step)

b2i(bytes1.first) < 0, ∀ n, bytes0. # bytes0 < #(bytes1 + by ') → b2i(bytes0.first) < 0 → bytes2int(mklist(i2b(~1), n) + bytes0) = bytes2int(bytes0) ∨ bytes0 = [] ⊦ bytes1 = [], mklist(i2b(~1), n) + bytes1 = [], bytes2int(mklist(i2b(~1), n) + bytes1) = bytes2int(bytes1)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [n, bytes0] → [n, 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: n, bytes1