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Proof step 26 from proof plinksd+ (previous step)

dnode(md, sdt)[p] = fnode(fid), p ∈ dnode(md, sdt) ⊦ str ∈ sdt, p ∈ dnode(md, sdt[str, tree]) ∧ dnode(md, sdt[str, tree])[p] = fnode(fid)

User applied rule constructor cut

And closed the goal!

Rule argument

var: p

Additional information

Used simplifier rules: path freely generated by ε, +;;% ⊦ str + p = str ' + p; (s)% ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ∈ sdt → dnode(md, sdt)[str ' + p] = sdt[str][p]; (s)% ⊦ str0 ∈ sdt ∧ ¬ str ∈ sdt → sdt[str, tree][str0] = sdt[str0]; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)% ⊦ tree[ε] = tree; (s)% ⊦ fnode(fid) ≠ dnode(md, sdt); (s)