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Proof step 22 from proof in-del (previous step)

p0 ∈ sdt[str], str ∈ sdt, p0 ∈ sdt[str0] → (q0 ∈ sdt[str0] -- p0 ↔ ¬ p0 ⊑ q0 ∧ q0 ∈ sdt[str0]), ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ⊦ str0 ∈ sdt[str, sdt[str] -- p0] ∧ q0 ∈ dnode(md, sdt[str, sdt[str] -- p0])[str0 '] ↔ ¬ str ' + p0 ⊑ str0 ' + q0 ∧ str0 ∈ sdt ∧ q0 ∈ sdt[str0], p0 = ε

Heuristic nonsplitting cut applied rule cut formula

And obtained the following new goals

Proof step 23:
str ∈ sdt, p0 ∈ sdt[str], str0 ∈ sdt[str, sdt[str] -- p0], p0 ∈ sdt[str0] → (q0 ∈ sdt[str0] -- p0 ↔ ¬ p0 ⊑ q0 ∧ q0 ∈ sdt[str0]), ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ⊦ str0 ∈ sdt[str, sdt[str] -- p0] ∧ q0 ∈ dnode(md, sdt[str, sdt[str] -- p0])[str0 '] ↔ ¬ str ' + p0 ⊑ str0 ' + q0 ∧ str0 ∈ sdt ∧ q0 ∈ sdt[str0], p0 = ε

Rule argument

fma: str0 ∈ sdt[str, sdt[str] -- p0]

Additional information

Used simplifier rules: ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws) Used formulas: str0 ∈ sdt[str, sdt[str] -- p0]