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Proof step 71 from proof bytes2bytes-zero (previous step)

b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes1. # bytes1 < #(bytes0 + by1 ') → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes1))) = by ' + by0 ' + bytes1 ⊦ byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0) = [], bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes0))) = by ' + by0 ' + bytes0, ¬ true ∈ byte2bits(by) ∧ ¬ true ∈ byte2bits(by0) ∧ ¬ true ∈ bytes2bits(bytes0), ¬ false ∈ byte2bits(by) ∧ ¬ false ∈ byte2bits(by0) ∧ ¬ false ∈ bytes2bits(bytes0)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [by0, by, bytes1] → [by0, by, bytes0]

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ #(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, by, bytes0