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Proof step 6 from proof subset-insert-02 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a1 ∈ s ++ a ++ a0, c ∈ s ⊦ c = a1, c = b, a0 = a, a ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c, a0 ∈ s
Proof step 15:
c ∈ s ⊦ a1 ∈ s ++ a ++ a0, c = a1, c = b, a0 = a, a ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c, a0 ∈ s

Rule argument

fma: a1 ∈ s ++ a ++ a0

Additional information

Used formulas: a1 ∈ s ++ a ++ a0