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Proof step 13 from proof fids-del (previous step)

p0 ∈ tree, p0 ∈ tree → (tree -- p0).fids = tree.fids \ tree[p0].fids ∨ p0 = ε ⊦ (dnode(md, sdt0[str, tree]) -- (str ' + p0)).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, str ∈ sdt0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ dnode(md, sdt).fids = sdt.fids
Proof step 15:
(tree -- p0).fids = tree.fids \ tree[p0].fids, p0 ∈ tree ⊦ (dnode(md, sdt0[str, tree]) -- (str ' + p0)).fids = (sdt0.fids ∪ tree.fids) \ tree[p0].fids, p0 = ε, str ∈ sdt0

Rule argument

Additional information

Used simplifier rules: fids-dnode: ⊦ dnode(md, sdt).fids = sdt.fids;%union-com: ⊦ fidb ∪ fidb0 = fidb0 ∪ fidb; from specification bag-union[fidbag]%union-ass: ⊦ fidb ∪ fidb0 ∪ fidb1 = (fidb ∪ fidb0) ∪ fidb1; from specification bag-union[fidbag]% ⊦ (fidb ∪ fidb0) \ fidb0 = fidb; (s)% ⊦ tree[ε] = tree; (s)% ⊦ ¬ str ∈ sdt → sdt -- str = sdt; (s)% ⊦ sdt[str, tree] -- str = sdt -- str; (s)% ⊦ dnode(md, sdt) -- str ' = dnode(md, sdt -- str); (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ p + ε = p; (s)