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Proof step 9 from proof bits2int-le8 (previous step)

# int2bits(bits2int(bits)) = i→n(i + 2), bits2int(bits) < i, bits2int(bits) < 2 ^ i, 0 < 2 ^ i, bits2int(bits) + 1 < 2 ^ i, bits2int(bits) < 2 ^ i * 2, bits2int(bits) + 1 < 2 ^ i * 2, 0 ≤ bits2int(bits) + 2 ^ i * 2, 0 ≤ i, bits2int(bits) + 2 ^ i < 0, # bits ≤ 8, bits2int(bits) < ~128, bits2int(bits) + 2 ^ i < i, bits2int(bits) < ~1, bits2int(bits) < 0 ⊦ bits = [], i = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# sgn(bits) = i→n(i + 2), bits2int(bits) < i, bits2int(bits) < 2 ^ i, 0 < 2 ^ i, bits2int(bits) + 1 < 2 ^ i, bits2int(bits) < 2 ^ i * 2, bits2int(bits) + 1 < 2 ^ i * 2, 0 ≤ bits2int(bits) + 2 ^ i * 2, 0 ≤ i, bits2int(bits) + 2 ^ i < 0, # bits ≤ 8, bits2int(bits) < ~128, bits2int(bits) + 2 ^ i < i, bits2int(bits) < ~1, bits2int(bits) < 0 ⊦ bits = [], i = ~1

Rule argument

Additional information

Used simplifier rules: bits2bits-def: bits ≠ [] ⊦ int2bits(bits2int(bits)) = sgn(bits);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;