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Proof step 8 from proof plinks-fids (previous step)

dnode(md, sdt) = tree0, ∀ str. str ∈ sdt → (∀ fid1. # links(fid1, sdt[str]) = count(fid1, sdt[str].fids)) ⊦ # links(fid1, dnode(md, sdt)) = count(fid1, sdt.fids)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ str. str ∈ sdt → (∀ fid1. # links(fid1, sdt[str]) = count(fid1, sdt[str].fids)) ⊦ # links(fid1, dnode(md, sdt)) = count(fid1, sdt.fids)

Proof step 8 from proof plinks-fids (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information