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Proof step 15 from proof subset-insert-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
c ∈ s ⊦ c = b, a0 = a, a1 = a0, a1 = a, a ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c, a0 ∈ s, a1 ∈ s

Rule argument

Additional information

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