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Proof step 16 from proof plinks-fids (previous step)

∀ str0. str0 ∈ sdt0[str, tree] → (∀ fid1. # links(fid1, sdt0[str, tree][str0]) = count(fid1, sdt0[str, tree][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 ∪ tree.fids), str ∈ sdt0

User applied rule all left

And obtained the following new goals

Proof step 17:
str ∈ sdt0[str, tree] → (∀ fid1. # links(fid1, sdt0[str, tree][str]) = count(fid1, sdt0[str, tree][str].fids)), ∀ str0. str0 ∈ sdt0[str, tree] → (∀ fid1. # links(fid1, sdt0[str, tree][str0]) = count(fid1, sdt0[str, tree][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 ∪ tree.fids), str ∈ sdt0

Rule argument

pos: left 1 terms: str abort: F discard: F computed: T

Additional information

Used terms: str