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Proof step 1 from proof int2bits-len-pos-suc (conclusion)

⊦ ¬(∃ i. 2 ^ i - 1 = j) ∧ 0 ≤ j → # int2bits(j) = # int2bits(j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ j ⊦ # int2bits(j) = # int2bits(j + 1), ∃ i. 2 ^ i = j + 1

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i - j = i0 ↔ i = i0 + j;