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Proof step 7 from proof int2uint-neg (previous step)

¬(0 < ~ j ∧ true ∈ uint2bits(~ j).rest.rest), j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(~ j)) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(~ j)) +1)) + j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j < 0 ⊦ bits2uint(sgn(bitsinvert(uint2bits(~ j)) +1)) = 2 ^ n→i(# sgn(bitsinvert(uint2bits(~ j)) +1)) + j, true ∈ uint2bits(~ j).rest.rest

Rule argument

Additional information

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