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Proof step 6 from proof plinksd+ (previous step)

⊦ p ∈ links(fid, dnode(md, sdt[str, tree])) ↔ p ∈ 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 7:
⊦ p ∈ ps ∪ ps0 ↔ p ∈ ps ∨ p ∈ ps0
Proof step 8:
⊦ p ∈ links(fid, dnode(md, sdt[str, tree])) ↔ p ∈ links(fid, dnode(md, sdt)) ∨ p ∈ str + links(fid, tree), str ∈ sdt

Rule argument

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

Additional information

Used terms: str + links(fid, tree), links(fid, dnode(md, sdt)), p