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Proof step 1 from proof int2bits-exneg (conclusion)

i < ~ 1 ⊦ ∃ bits. int2bits(i) = true ' + false ' + bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i < ~1, i < 0 ⊦ ∃ bits. int2bits(i) = true ' + false ' + bits

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)