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Proof step 14 from proof int2bits-lelen-07 (previous step)

j < 0 ∧ 0 ≤ n→i(# int2bits(~ j)) ∧ ~ j ≠ 2 ^ n→i(# int2bits(~ j)), ¬ # int2bits(j) ≤ i→n(n→i(# int2bits(~ j))) + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j < 0, 1 + # int2bits(~ j) < # int2bits(j) ⊦ j + 2 ^ n→i(# int2bits(~ j)) = 0

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i→n(n→i(n)) = n;% ⊦ 0 ≤ n→i(n);% ⊦ ~ i = j ↔ i + j = 0;