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Proof step 49 from proof int2bytes-rec-pos (previous step)

¬(uint2bits(i % 256) ≠ [] ∧ # uint2bits(i % 256) < 9), uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, # uint2bits(i % 256) ≤ 9, 255 < i, 0 < i, 0 ≤ i, i % 256 ≤ 127 ⊦ fillsgn(9, uint2bits(i % 256)) = [], i2b(bits2int(fillsgn(9, uint2bits(i % 256)).rest)) = i2b(i % 256), bits2int(fillsgn(9, uint2bits(i % 256)).rest) = i % 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
# uint2bits(i % 256) = 9, uint2bits(i) = uint2bits(i / 256) + fillsgn(9, uint2bits(i % 256)).rest, i % 256 ≤ 127, 0 ≤ i, 0 < i, 255 < i ⊦ bits2int(fillsgn(9, uint2bits(i % 256)).rest) = i % 256, i2b(bits2int(fillsgn(9, uint2bits(i % 256)).rest)) = i2b(i % 256), fillsgn(9, uint2bits(i % 256)) = []

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n ≤ m ↔ m = n);% ⊦ ¬ m < n ↔ n ≤ m;%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;