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Proof step 83 from proof bytes2bytes-mone (previous step)

bytes0 = by0 + bytes, 8 = # fillsgn(8, int2bits(b2i(by))), b2i(by0) = ~1, # bytes0 = 1, 0 ≤ b2i((bytes + 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 84:
b2i(by0) = ~1, 8 = # fillsgn(8, int2bits(b2i(by))), true ∈ byte2bits(by0), false ∈ byte2bits(by0), 0 ≤ b2i(by), ∀ bytes1. # bytes1 < #((by0 + []) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(byte2bits(by0))) = by0 '

Rule argument

Additional information

Used simplifier rules: bytes2bits-append: ⊦ bytes2bits(bytes + bytes0) = bytes2bits(bytes) + bytes2bits(bytes0);%bytes2bits-one: ⊦ bytes2bits(by ') = byte2bits(by);%bytes2bits-empty: ⊦ bytes2bits([]) = [];% ⊦ ¬ boolvar ∈ [];% ⊦ by '.first = by;% ⊦ [] + bytes = bytes;% ⊦ bytes + [] = bytes;% ⊦ bits + [] = bits;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ by ' + bytes ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ by + bytes = by ' + bytes;