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Proof step 8 from proof subset-insert-02 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 9:
a1 ∈ s, c ∈ s, b ∈ s ++ a ++ a0 ++ a1 ⊦ a0 = a, c = b, c = a1, a ∈ s, a0 ∈ s, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c
Proof step 12:
a1 ∈ s, c ∈ s ⊦ a0 = a, c = b, c = a1, a ∈ s, a0 ∈ s, b ∈ s ++ a ++ a0 ++ a1, s ⊆ s ++ a ++ a0 ++ a1 ++ b ++ c

Rule argument

fma: b ∈ s ++ a ++ a0 ++ a1

Additional information

Used formulas: b ∈ s ++ a ++ a0 ++ a1