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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);