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Proof step 250 from proof int2bytes-rec-pos-any (previous step)

#(mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256)) ≤ 8, 0 ≤ j, 0 < j, 256 ≤ j, # int2bits(j % 256) < 8, 0 ≤ b2i(i2b(j % 256)) ⊦ mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256) = [], 9 = # int2bits(j % 256), # int2bits(j % 256) = 8, i2b(bits2int(mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256))) ' = i2b(j % 256) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n < m → (m - n) + n = m;% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ by ' = by0 ' ↔ by = by0;