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Proof step 4 from proof subset-insert-02 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
c ∈ s ⊦ c = a0, c = a1, c = b, a ∈ s, a0 ∈ s ++ a, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c

Rule argument

fma: a0 ∈ s ++ a

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%add-01: ⊦ s ++ a ++ a = s ++ a;%subset-insert-01: ⊦ s ⊆ s ++ a ++ a0 ++ b ++ c;%ins-two: ⊦ a ∈ s → s ++ b ++ a = s ++ b; Used formulas: a0 ∈ s ++ a