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Proof step 13 from proof again-put (previous step)

str ∈ sdt, p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → tree[p0, tree[p0]] = tree ∨ p0 = ε ⊦ dnode(md, sdt[str, sdt[str][p0, sdt[str][p0]]]) = dnode(md, sdt)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 14:
⊦ tree[ε] = tree
Proof step 15:
⊦ tree[ε, tree0] = tree0
Proof step 16:
p0 ∈ sdt[str], str ∈ sdt, ∀ tree. p0 ∈ tree → tree[p0, tree[p0]] = tree ∨ p0 = ε ⊦ dnode(md, sdt[str, sdt[str][p0, sdt[str][p0]]]) = dnode(md, sdt), p0 = ε

Rule argument

fma: p0 = ε

Additional information

Used simplifier rules: nil-at: ⊦ tree[ε] = tree;%nil-put: ⊦ tree[ε, tree0] = tree0;% ⊦ str ∈ sdt → sdt[str, sdt[str]] = sdt; (s) Used formulas: p0 = ε