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Proof step 108 from proof byte-law-righthift-not (previous step)

0 ≤ 128, i2b(bits2uint(firstn(8, uint2bits(128)))) = i2b(2147483392 + bits2uint(true ' + firstn(7, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))), # uint2bits(128) ≤ 32 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 109:
i2b(bits2uint(firstn(8, (((((((false ' + true ') + false ') + false ') + false ') + false ') + false ') + false ') + false '))) = i2b(2147483392 + bits2uint(true ' + firstn(7, true ' + false ' + false ' + false ' + false ' + false ' + false ' + false '))) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);