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Proof step 17 from proof subset-insert-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
c ∈ s, b ∈ s ⊦ a1 = a, a1 = a0, a0 = a, c = b, a1 ∈ s, a0 ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c, a ∈ s

Rule argument

Additional information

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