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Proof step 12 from proof at-put (previous step)

q ∈ dnode(md, sdt[str, tree0]) ⊦ dnode(md, sdt[str, tree0])[q] = (str ' ⊑ q ⊃ tree0[q.tail];(q = ε ⊃ dnode(md, sdt[str, tree0]);dnode(md, sdt)[q]))

User applied rule constructor cut

And obtained the following new goals

Proof step 13:
⊦ tree[ε] = tree
Proof step 14:
q = str0 + q0, q ∈ dnode(md, sdt[str, tree0]) ⊦ dnode(md, sdt[str, tree0])[q] = (str ' ⊑ q ⊃ tree0[q.tail];(q = ε ⊃ dnode(md, sdt[str, tree0]);dnode(md, sdt)[q]))

Rule argument

var: q

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%path freely generated by ε, +;;% ⊦ p ⊑ ε ↔ p = ε; (s)% ⊦ str ' ≠ ε; (s)