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Proof step 3 from proof subset (previous step)

b ∈ s ⊦ b = a, s ⊆ s ++ a ++ b

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
b ∈ s ⊦ b = a, a ∈ s, s ⊆ s ++ a ++ b

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%sub-idem: ⊦ s ⊆ s; Used formulas: a ∈ s