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Proof step 135 from proof byte-law-righthift-not (previous step)

0 ≤ 2147483584, i2b(64) = i2b(bits2int(lastn(8, uint2bits(2147483584)))) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 136:
i2b(64) = i2b(bits2int(((((((true ' + true ') + false ') + false ') + false ') + false ') + false ') + false ')) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (boolvar ' + boolvar0 ' + bits).last ↔ (boolvar0 ' + bits).last;% ⊦ lastn(1, bits) = bits.last ';%0 < n ⊦ lastn((* n), bits + boolvar ') = lastn(n -1, bits) + boolvar ';% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);