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Proof step 72 from proof bytes2bytes-mone (previous step)

8 = # fillsgn(8, int2bits(b2i(by))), b2i(bytes0.first) = ~1, 0 ≤ b2i((bytes0.rest + by ').first), false ∈ bytes2bits(bytes0), true ∈ bytes2bits(bytes0), ∀ bytes1. # bytes1 < #(bytes0 + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bytes0 = [], bits2bytes(sgn(bytes2bits(bytes0))) = bytes0

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [bytes1] → [bytes0]

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: bytes0