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Proof step 12 from proof subset-insert-02 (previous step)

a1 ∈ s, c ∈ s ⊦ a0 = a, c = b, c = a1, a ∈ s, a0 ∈ s, b ∈ s ++ a ++ a0 ++ a1, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
c ∈ s, a1 ∈ s ⊦ a0 = a, b = a0, b = a, c = a1, a0 ∈ s, a ∈ s, b ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;