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

c ∈ s ⊦ c = a, c = b, s ⊆ s ++ a ++ b ++ c

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%ins-two: ⊦ a ∈ s → s ++ b ++ a = s ++ b;%sub-add-01: ⊦ s ⊆ s ++ a; Used formulas: a ∈ s