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Proof step 5 from proof fids-cons-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid)
Proof step 7:
root[p] = fnode(fid), p ∈ root, ∀ fid0. fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid0) ⊦ fid ∈ fs

Rule argument

Additional information

Used simplifier rules: ofh-fids: ⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid);