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Proof step 35 from proof plinks-fids (previous step)

# links(fid1, dnode(md, sdt0)) = count(fid1, sdt0.fids), dnode(md, sdt0)[p] = fnode(fid1), tree[p.tail] = fnode(fid1), p ∈ dnode(md, sdt0), p.tail ∈ tree, ∀ str0. str0 ∈ sdt0[p.head, tree] → (∀ fid1. # links(fid1, sdt0[p.head, tree][str0]) = count(fid1, sdt0[p.head, tree][str0].fids)), ∀ sdt1, fid1, md. # sdt1 < # sdt0[p.head, 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)) ∩ (p.head + links(fid1, tree))) = 0, #(links(fid1, dnode(md, sdt0)) ∪ (p.head + links(fid1, tree))) = count(fid1, sdt0.fids) + # links(fid1, tree), p = ε, links(fid1, dnode(md, sdt0)) ∩ (p.head + links(fid1, tree)) = ∅, p.head ∈ sdt0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data, inst: path name: elim-list-c seq: p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0 subst: [p, str, p0] → [p, str1, p0]

Additional information

Used simplifier rules: ⊦ str + p = str ' + p; (s)% ⊦ str ' + p ≠ ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ' + p ∈ dnode(md, sdt) ↔ str ∈ sdt ∧ p ∈ sdt[str]; (s)%p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0; (elim) Used terms: p, str1, p0