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Proof step 4 from proof subset-insert-01 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: a0 ∈ s ++ a

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;%ins-two: ⊦ a ∈ s → s ++ b ++ a = s ++ b; Used formulas: a0 ∈ s ++ a