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Proof step 9 from proof subset-insert-02 (previous step)

b ∈ s ++ a ++ a0 ++ a1, c ∈ s, a1 ∈ s ⊦ 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 10:
a1 ∈ s, c ∈ s, b ∈ s ⊦ c = a1, c = b, a0 = a, a ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c, a0 ∈ s

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%addtwo: ⊦ s ++ a ++ b ++ a = s ++ b ++ a;%ins-two: ⊦ a ∈ s → s ++ b ++ a = s ++ b;%subset-insert: ⊦ s ⊆ s ++ a ++ b ++ c;%addthree: ⊦ s ++ a ++ b ++ c ++ a = s ++ a ++ b ++ c;%subset-insert-01: ⊦ s ⊆ s ++ a ++ a0 ++ b ++ c;