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Proof step 5 from proof plinks-fids (previous step)

⊦ (fnode(fid) = tree → # (fid0 = fid ⊃ ∅ ++ ε;∅) = (fid0 = fid ⊃ 1;0)) ∧ ( (∀ str. str ∈ sdt → (∀ fid1. # links(fid1, sdt[str]) = count(fid1, sdt[str].fids))) → dnode(md, sdt) = tree0 → # links(fid1, dnode(md, sdt)) = count(fid1, sdt.fids))

User applied rule case distinction

And obtained the following new goals

Proof step 6:
fnode(fid) = tree ⊦ # (fid0 = fid ⊃ ∅ ++ ε;∅) = (fid0 = fid ⊃ 1;0)
Proof step 8:
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)

Rule argument

pos: right 1

Proof step 5 from proof plinks-fids (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information