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Proof step 4 from proof subset-insert (previous step)

c ∈ s ⊦ a ∈ 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 5:
c ∈ s ⊦ b ∈ s ++ a, a ∈ s, c = a, c = b, s ⊆ s ++ a ++ b ++ c

Rule argument

fma: b ∈ s ++ a

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%add-01: ⊦ s ++ a ++ a = s ++ a;%ins-two: ⊦ a ∈ s → s ++ b ++ a = s ++ b;%sub-insert: ⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0);%sub-idem: ⊦ s ⊆ s; Used formulas: b ∈ s ++ a