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Proof step 30 from proof bytes2bytes-mone (previous step)

8 < #(true ' + byte2bits(by)), b2i(by0) = ~1, 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(firstn(#(true ' + byte2bits(by)) - 8, true ' + byte2bits(by))) + i2b(bits2int(lastn(8, true ' + byte2bits(by)))) = by0 ' + by '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: byte2bits-length: ⊦ # byte2bits(by) = 8;% ⊦ bytes + by = bytes + by ';% ⊦ firstn(1, boolvar ' + bits) = boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;