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Proof step 2 from proof subset-insert-01 (previous step)

c ∈ s ++ a ++ a0 ++ b ⊦ s ⊆ s ++ a ++ a0 ++ b ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
c ∈ s ⊦ c = a, c = a0, c = b, s ⊆ s ++ a ++ a0 ++ b ++ c

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%add-01: ⊦ s ++ a ++ a = s ++ a;%subset-insert: ⊦ s ⊆ s ++ a ++ b ++ c;%addtwo: ⊦ s ++ a ++ b ++ a = s ++ b ++ a;%addthree: ⊦ s ++ a ++ b ++ c ++ a = s ++ a ++ b ++ c;%sub-insert: ⊦ ¬ a ∈ s → (s ⊆ s0 ++ a ↔ s ⊆ s0);%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);