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Proof step 1 from proof int2bits-2pot-03 (conclusion)

0 ≤ j, ¬ false ∈ int2bits(j).rest ⊦ j = 2 ^ n→i(# int2bits(j).rest) - 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ j ⊦ 2 ^ n→i(# int2bits(j).rest) = j + 1, false ∈ int2bits(j).rest

Rule argument

Additional information

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