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Proof step 6 from proof sub-insert (previous step)

s ⊆ s0 ++ a ⊦ a ∈ s, a ∈ s0, s ⊆ s0, s0 ++ a ⊆ s0

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
s ⊆ s0 ++ a, s0 ⊆ s0 ⊦ a ∈ s0, a ∈ s, s0 ++ a ⊆ s0, s ⊆ s0

Rule argument

fma: s0 ⊆ s0

Additional information

Used simplifier rules: sub-idem: ⊦ s ⊆ s; Used formulas: s0 ⊆ s0