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Proof step 15 from proof int2bits-neg-03 (previous step)

¬ 0 ≤ 2 ^ n→i(# int2bits(~ j)) + j, int2bits(2 ^ n→i(# int2bits(~ j)) + j) = boolvar ', j < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
int2bits(2 ^ n→i(# int2bits(~ j)) + j) = boolvar ', j < 0, 2 ^ n→i(# int2bits(~ j)) + j < 0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;