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Proof step 102 from proof byte-law-righthift-not (previous step)

#(true ' + firstn(n0, int2bits(~128))) ≠ 0, i2b(bits2uint(firstn(# int2bits(~128), int2bits(128)))) = i2b(bits2uint(mklist(true, n)) * 2 ^ n→i(#(true ' + firstn(n0, int2bits(~128)))) + bits2uint(true ' + firstn(n0, int2bits(~128)))), 31 = # int2bits(~128) + n, 32 = n + # int2bits(128), 30 = n0 + n, n ≤ 31, # int2bits(128) ≤ 32, # int2bits(~128) ≤ 32, n ≤ 32, n ≤ 30 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # firstn(n, bits) = n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;