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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%negle-13: ⊦ ~ i ≤ j ↔ 0 ≤ i + j; from specification int-abs