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Proof step 9 from proof in-del (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 10:
⊦ ε ∈ tree
Proof step 11:
q = str0 + q0, p0 ∈ sdt[str], str ∈ sdt, ∀ tree, q. p0 ∈ tree → (q ∈ tree -- p0 ↔ ¬ p0 ⊑ q ∧ q ∈ tree) ∨ p0 = ε ⊦ q ∈ dnode(md, sdt) -- (str ' + p0) ↔ ¬ str ' + p0 ⊑ q ∧ q ∈ dnode(md, sdt)

Rule argument

var: q

Additional information

Used simplifier rules: nil-in: ⊦ ε ∈ tree;%path freely generated by ε, +;;% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ str ' + p ≠ ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]