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Proof step 30 from proof inv-close (previous step)

ofh0[fd] = mkfh(fid, mode, m), ofh0[n] = mkfh(fid0, mode0, m0), root.dir?, fids-cons(root, fs, ofh0), fid0 ∈ fs, fd ∈ ofh0, n ∈ ofh0 ⊦ # ofh0.fids = 0, ofh0.fids = ∅, fid0 = fid, fid ∈ root.fids, fid ∈ (ofh0 -- fd).fids, fid0 ∈ root.fids, ∃ n0. n0 ≠ fd ∧ n0 ∈ ofh0 ∧ ofh0[n0].fid = fid0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 7 terms: n abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ mkfh(fid, mode, m).fid = fid; (s)%⊦ mkfh(fid, mode, m) = mkfh(fid0, mode0, m0) ↔ fid = fid0 ∧ mode = mode0 ∧ m = m0; (ws) Used terms: n