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Proof step 5 from proof int2bits-leneq-neg (previous step)

j < 0 ⊦ ¬ 2 ^ (n→i(# int2bits(j)) - 1) + j < 0, # int2bits(j) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j < 0, 2 ^ (n→i(# int2bits(j)) - 1) + j < 0 ⊦ # int2bits(j) = 1

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs