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Proof step 8 from proof int2bytes-minus-256 (previous step)

8 < #(true ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ') ⊦ bits2bytes (firstn (#(true ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ') - 8, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false ')) + i2b(bits2int(lastn(8, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))) = i2b(~1) ' + i2b(0) '

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: bits2bytes-true: ⊦ bits2bytes(true ') = i2b(~1) ';% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ bytes + by = bytes + by ';% ⊦ bits2uint(false ') = 0;% ⊦ bits2uint(bits + false ') = bits2uint(bits) * 2;% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ lastn(1, bits) = bits.last ';%0 < n ⊦ lastn((* n), bits + boolvar ') = lastn(n -1, bits) + boolvar ';% ⊦ firstn(1, boolvar ' + bits) = boolvar ';% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;