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Proof step 5 from proof subset-insert-01 (previous step)

c ∈ s ⊦ a0 ∈ s ++ a, a ∈ s, c = a, c = a0, c = b, s ⊆ s ++ a ++ a0 ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
c ∈ s ⊦ c = b, a0 = a, a ∈ s, s ⊆ s ++ a ++ a0 ++ b ++ c, a0 ∈ s

Rule argument

Additional information

Used simplifier rules: add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;