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Proof step 10 from proof plinks-fids (previous step)

∀ str. str ∈ sdt → (∀ fid1. # links(fid1, sdt[str]) = count(fid1, sdt[str].fids)), ∀ sdt0, fid1, md. # sdt0 < # sdt → (∀ str. str ∈ sdt0 → (∀ fid1. # links(fid1, sdt0[str]) = count(fid1, sdt0[str].fids))) → # links(fid1, dnode(md, sdt0)) = count(fid1, sdt0.fids) ⊦ # links(fid1, dnode(md, sdt)) = count(fid1, sdt.fids)

User applied rule cut formula

And obtained the following new goals

Proof step 11:
⊦ links(fid, dnode(md, ∅)) = ∅
Proof step 12:
∀ str. str ∈ sdt → (∀ fid1. # links(fid1, sdt[str]) = count(fid1, sdt[str].fids)), ∀ sdt0, fid1, md. # sdt0 < # sdt → (∀ str. str ∈ sdt0 → (∀ fid1. # links(fid1, sdt0[str]) = count(fid1, sdt0[str].fids))) → # links(fid1, dnode(md, sdt0)) = count(fid1, sdt0.fids) ⊦ # links(fid1, dnode(md, sdt)) = count(fid1, sdt.fids), sdt = ∅

Rule argument

fma: sdt = ∅

Additional information

Used simplifier rules: linksd0: ⊦ links(fid, dnode(md, ∅)) = ∅;% ⊦ # ∅ = 0; (s)% ⊦ ∅.fids = ∅; (s)% ⊦ count(fid, ∅) = 0; (s) Used formulas: sdt = ∅