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Proof step 19 from proof fids-sub (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic[directory-tree-inst]%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat[directory-tree-inst]