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Proof step 4 from proof int2bits-lelen-07 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 ^ n→i(# int2bits(~ j)) + j ≤ 0 ⊦ ¬ 0 ≤ j ∧ j + 2 ^ n→i(# int2bits(~ j)) ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0;% ⊦ 0 ≤ n→i(n);% ⊦ ¬ i < j ↔ j ≤ i;