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Proof step 7 from proof ofh-fids-new (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 8:
∃ n. n ∈ ofh[fd, fh] ∧ ofh[fd, fh][n].fid = fid ⊦ fid = fh.fid, fd ∈ ofh, ∃ n. n ∈ ofh ∧ ofh[n].fid = fid
Proof step 14:
fid = fh.fid ∨ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid) ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, fh] ∧ ofh[fd, fh][n].fid = fid

Rule argument

pos: right 1

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