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Proof step 3 from proof sub-insert (previous step)

⊦ s ⊆ s0 ++ a ↔ s ⊆ s0, a ∈ s

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
⊦ s ⊆ s0 ++ a ↔ s ⊆ s0, a ∈ s, a ∈ s0

Rule argument

fma: a ∈ s0

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s; Used formulas: a ∈ s0