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Proof step 64 from proof int2bytes-rec-pos (previous step)

9 ≤ # uint2bits(i % 256), uint2bits(i) = uint2bits(i / 256) + uint2bits(i % 256).rest, i % 256 < i, 0 < i % 256, 0 ≤ i, 255 < i, # uint2bits(i % 256) ≤ 9, i % 256 ≤ 255, 127 < i % 256, 0 < i, 1 < i, 127 < i, 128 < i ⊦ i2b(bits2int(uint2bits(i % 256).rest)) = i2b(i % 256 - 256), uint2bits(i % 256) = [], i % 256 = bits2int(uint2bits(i % 256).rest) + 256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 65:
# uint2bits(i % 256) = 9, uint2bits(i) = uint2bits(i / 256) + uint2bits(i % 256).rest, 128 < i, 127 < i, 1 < i, 0 < i, 127 < i % 256, i % 256 ≤ 255, 255 < i, 0 ≤ i, 0 < i % 256, i % 256 < i ⊦ i % 256 = bits2int(uint2bits(i % 256).rest) + 256, i2b(bits2int(uint2bits(i % 256).rest)) = i2b(i % 256 - 256)

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;% ⊦ i2b(i) ≠ i2b(j) → i ≠ j; (forward)% ⊦ m ≤ n → (n ≤ m ↔ m = n);%0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;