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Proof step 247 from proof int2bytes-rec-pos-any (previous step)

# int2bits(j % 256) ≤ 8, 0 ≤ bytes2int(i2b(bits2int(int2bits(j % 256))) '), 0 ≤ j, 0 < j, 256 ≤ j, # int2bits(j % 256) < 8 ⊦ bits2bytes(mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256)) = i2b(bits2int(int2bits(j % 256))) ', mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256) = [], 9 = # int2bits(j % 256), # int2bits(j % 256) = 8

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 248:
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, bits2bytes(mklist(false, 8 - # int2bits(j % 256)) + int2bits(j % 256)) = i2b(j % 256) '

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ m ≠ n → (m ≤ n ↔ m < n);% ⊦ bits2int(int2bits(i)) = i;