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Proof step 13 from proof inv-close (previous step)

ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fd ∈ ofh0 ⊦ fids-cons(root, fs -- fid, ofh0 -- fd), fid ∈ (ofh0 -- fd).fids, fid ∈ root.fids

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ fids-cons(tree0, fs0, ofh0) → (fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)))
Proof step 15:
fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ fid0 ∈ ofh0.fids, ofh0[fd] = mkfh(fid, mode, m), root.dir?, fids-cons(root, fs, ofh0), fd ∈ ofh0 ⊦ fid0 ≠ fid ∧ fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ fid0 ∈ (ofh0 -- fd).fids, fid ∈ root.fids, fid ∈ (ofh0 -- fd).fids

Rule argument

Additional information

Used simplifier rules: fids-cons-reduce: ⊦ fids-cons(tree0, fs0, ofh0) → ( fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)));% ⊦ fid ∈ fs -- fid0 ↔ fid ≠ fid0 ∧ fid ∈ fs; (s)