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Proof step 18 from proof fids-sub (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 19:
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

Rule argument

pos: left 3 terms: fid abort: F discard: F computed: T

Additional information

Used terms: fid