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Proof step 3 from proof one-in (previous step)

# s = 1, a ∈ s ⊦ s = ∅ ++ a

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
# s = 1, a ∈ s ⊦ s = ∅ ++ a, a ∈ ∅

Rule argument

fma: a ∈ ∅

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%In-empty: ⊦ ¬ a ∈ ∅; Used formulas: a ∈ ∅