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Proof step 21 from proof bytes2bytes-first-neg (previous step)

b2i(int2bytes(bytes2int(by0 ' + bytes1)).first) < 0, 0 ≤ b2i(by0), ∀ by, bytes0. # bytes0 < #(by0 + bytes1) → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [by, bytes0] → [by0, bytes1]

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ by + bytes = by ' + bytes;% ⊦ #(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: by0, bytes1