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Proof step 3 from proof in-dnode-empty (previous step)

p ∈ dnode(md, ∅) ⊦ p = ε

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]

Rule argument

var: p

Additional information

Used simplifier rules: in-dnode: ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str];%path freely generated by ε, +;;% ⊦ str + p = str ' + p; (s)% ⊦ str ' + p ≠ ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ ¬ str ∈ ∅; (s)