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Proof step 20 from proof fids-sub (previous step)

count(fid, sdt0.fids ∪ tree.fids) < count(fid, tree[p0].fids), count(fid, sdt0.fids ∪ tree.fids) < count(fid, tree.fids), p0 ∈ tree, count(fid, tree[p0].fids) ≤ count(fid, tree.fids), ∀ fid. count(fid, tree[p0].fids) ≤ count(fid, tree.fids) ⊦ str ∈ sdt0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: bag-union, inst: fidbag name: union seq: ⊦ count(fid, fidb ∪ fidb0) = count(fid, fidb) + count(fid, fidb0) subst: [fidb0, fidb, fid] → [tree.fids, sdt0.fids, fid] flags: <all paths>

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat[directory-tree-inst]% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic[directory-tree-inst]%c: ⊦ m + n = n + m; from specification nat-basic[directory-tree-inst] Used terms: tree.fids, sdt0.fids, fid