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Proof step 5 from proof bint-unsigned (previous step)

bi2i(i2bi(i)) ≠ i - 4294967296, 0 ≤ i, 0 < i, i < 4294967296, 2147483648 ≤ i ⊦ i2bi(i - 4294967296) = i2bi(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ i, 0 < i, i < 4294967296, 2147483648 ≤ i ⊦ i2bi(i - 4294967296) = i2bi(i), i = bi2i(i2bi(i)) + 4294967296

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;