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Proof step 3 from proof plinksd+ (previous step)

⊦ links(fid, dnode(md, sdt[str, tree])) = links(fid, dnode(md, sdt)) ∪ (str + links(fid, tree)), str ∈ sdt

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ps = ps0 ↔ (∀ p. p ∈ ps ↔ p ∈ ps0)
Proof step 5:
⊦ str ∈ sdt, ∀ p. p ∈ links(fid, dnode(md, sdt[str, tree])) ↔ p ∈ links(fid, dnode(md, sdt)) ∪ (str + links(fid, tree))

Rule argument

spec: set-basic, inst: pathset name: Extension seq: ⊦ ps = ps0 ↔ (∀ p. p ∈ ps ↔ p ∈ ps0) subst: [ps0, ps] → [links(fid, dnode(md, sdt)) ∪ (str + links(fid, tree)), links(fid, dnode(md, sdt[str, tree]))] flags: <all paths>

Additional information

Used terms: links(fid, dnode(md, sdt)) ∪ (str + links(fid, tree)), links(fid, dnode(md, sdt[str, tree]))