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Proof step 21 from proof bytes2int-len-one-01 (previous step)

bytes0 = by0 + bytes, b2i(by) < 128, 0 ≤ b2i(by) ⊦ bytes0 = [], by0 = i2b(0), true ∈ bytes2bits(bytes0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
b2i(by) < 128, 0 ≤ b2i(by) ⊦ by0 = i2b(0), true ∈ bytes2bits(bytes), true ∈ byte2bits(by0)

Rule argument

Additional information

Used simplifier rules: bytes2bits-rec: ⊦ bytes2bits(by ' + bytes) = byte2bits(by) + bytes2bits(bytes);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes ≠ [];% ⊦ by + bytes = by ' + bytes;