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Proof step 38 from proof inv-close (previous step)

ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fd ∈ ofh0, ∃ n. n ∈ ofh0 -- fd ∧ (ofh0 -- fd)[n].fid = fid0 ⊦ # (ofh0 -- fd).fids = 0, (ofh0 -- fd).fids = ∅, fid0 ∈ fs, fid0 ∈ ofh0.fids, fid0 ∈ root.fids, ¬ fid ∈ (ofh0 -- fd).fids ∧ ¬ fid ∈ root.fids

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fd ∈ ofh0, n ∈ ofh0 ⊦ # (ofh0 -- fd).fids = 0, (ofh0 -- fd).fids = ∅, n = fd, ofh0[n].fid ∈ root.fids, ofh0[n].fid ∈ ofh0.fids, ofh0[n].fid ∈ fs, ¬ fid ∈ (ofh0 -- fd).fids ∧ ¬ fid ∈ root.fids

Rule argument

Additional information

Used simplifier rules: ⊦ fd ≠ fd0 → (ofh -- fd0)[fd] = ofh[fd]; (s)% ⊦ fd ∈ ofh -- fd0 ↔ fd ≠ fd0 ∧ fd ∈ ofh; (s)