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Proof step 51 from proof inv-close (previous step)

ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fid ∈ (ofh0 -- fd).fids, fd ∈ ofh0, fid ∈ fs ⊦ # ofh0.fids = 0, # (ofh0 -- fd).fids = 0, (ofh0 -- fd).fids = ∅, ofh0.fids = ∅, fid ∈ root.fids, ∃ n0. n0 ≠ fd ∧ n0 ∈ ofh0 ∧ ofh0[n0].fid = fid

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 52:
⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid)
Proof step 53:
ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fid ∈ fs, fd ∈ ofh0, ∃ n. n ∈ ofh0 -- fd ∧ (ofh0 -- fd)[n].fid = fid ⊦ # (ofh0 -- fd).fids = 0, # ofh0.fids = 0, ofh0.fids = ∅, (ofh0 -- fd).fids = ∅, fid ∈ root.fids, ∃ n0. n0 ≠ fd ∧ n0 ∈ ofh0 ∧ ofh0[n0].fid = fid

Rule argument

spec: posix-base name: ofh-fids seq: ⊦ fid ∈ ofh.fids ↔ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid) subst: [ofh, fid] → [ofh0 -- fd, fid] flags: <all paths>

Additional information

Used terms: ofh0 -- fd, fid