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Proof step 1 from proof int2bits-leneq-pos (conclusion)

⊦ 0 ≤ j ∧ 1 < n → (# int2bits(j) = n ↔ 2 ^ (n→i(n) - 2) ≤ j ∧ j < 2 ^ (n→i(n) - 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ j ⊦ # int2bits(j) = n ↔ 2 ^ (n→i(n) - 2) ≤ j ∧ j < 2 ^ (n→i(n) - 1), n = 1, n = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1