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Proof step 5 from proof ofh-fids-new (previous step)

⊦ fd ∈ ofh, ∀ fid. fid ∈ ofh[fd, fh].fids ↔ fid ∈ ofh.fids ++ fh.fid

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:
⊦ (∃ n. n ∈ ofh[fd, fh] ∧ ofh[fd, fh][n].fid = fid) ↔ fid = fh.fid ∨ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid), fd ∈ ofh

Rule argument

Additional information

Used simplifier rules: ofh-fids: ⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid);% ⊦ fid ∈ fids ++ fid0 ↔ fid = fid0 ∨ fid ∈ fids; (s)