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Proof step 7 from proof fids-cons-in (previous step)

root[p] = fnode(fid), p ∈ root, ∀ fid0. fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid0) ⊦ fid ∈ fs

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: fid

Additional information

Used simplifier rules: ⊦ tree[p] = fnode(fid) ∧ p ∈ tree → fid ∈ tree.fids; (ss) Used terms: fid