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Proof step 48 from proof bytes2bytes-other (previous step)

∀ bytes1. # bytes1 < #((by0 + bytes) + by ') → bytes2bits(bytes1) ≠ [] → bytes1 = [] ∨ b2i(bytes1.first) = 0 ∨ b2i(bytes1.first) = ~1 ∨ bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ⊦ b2i(by0) = ~1, b2i(by0) = 0, bits2bytes(sgn(byte2bits(by0) + bytes2bits(bytes))) = by0 ' + bytes, ¬ false ∈ byte2bits(by0) ∧ ¬ false ∈ bytes2bits(bytes), ¬ true ∈ byte2bits(by0) ∧ ¬ true ∈ bytes2bits(bytes)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [bytes1] → [by0 ' + bytes]

Additional information

Used simplifier rules: bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);%byte2bits-empty: ⊦ byte2bits(by) ≠ [];%bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);% ⊦ (by ' + bytes).first = by;% ⊦ by ' ≠ [];% ⊦ ¬ bytes + bytes0 ≠ [] ↔ bytes = [] ∧ bytes0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ #(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