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Proof step 17 from proof at-put (previous step)

str0 ∈ sdt[str, tree0], q0 ∈ sdt[str, tree0][str0] ⊦ sdt[str, tree0][str0][q0] = (str = str0 ⊃ tree0[q0];dnode(md, sdt)[str0 ' + q0])

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Rule argument

fma: str0 = str

Additional information

Used simplifier rules: at-dnode: ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p];% ⊦ str ≠ str0 → sdt[str0, tree][str] = sdt[str]; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ sdt[str, tree][str] = tree; (s) Used formulas: str0 = str