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Proof step 2 from proof exnotempty (previous step)

⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
st = ∅ ⊦ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 6:
¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0) ⊦ st = ∅

Rule argument

pos: right 1

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