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Proof step 26 from proof plinks-fids (previous step)

str0 ∈ sdt0, str0 ∈ sdt0[str, tree], ∀ str0. str0 ∈ sdt0[str, tree] → (∀ fid1. # links(fid1, sdt0[str, tree][str0]) = count(fid1, sdt0[str, tree][str0].fids)), ∀ fid0. # links(fid0, sdt0[str0]) = count(fid0, sdt0[str0].fids), ∀ sdt1, fid1, md. # sdt1 < # sdt0[str, tree] → (∀ str. str ∈ sdt1 → (∀ fid1. # links(fid1, sdt1[str]) = count(fid1, sdt1[str].fids))) → # links(fid1, dnode(md, sdt1)) = count(fid1, sdt1.fids) ⊦ #(links(fid1, dnode(md, sdt0)) ∪ (str + links(fid1, tree))) = count(fid1, sdt0.fids) + # links(fid1, tree), # links(fid, sdt0[str0]) = count(fid, sdt0[str0].fids), str ∈ sdt0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: fid

Additional information

Used terms: fid