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Proof step 1 from proof subset-insert (conclusion)

⊦ s ⊆ s ++ a ++ b ++ c

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 2:
c ∈ s ++ a ++ b ⊦ s ⊆ s ++ a ++ b ++ c

Rule argument

fma: c ∈ s ++ a ++ b

Additional information

Used simplifier rules: add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%sub-insert: ⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0);%subset: ⊦ s ⊆ s ++ a ++ b; Used formulas: c ∈ s ++ a ++ b