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Proof step 2 from proof fid-in-fs (previous step)

ofh[fd] = mkfh(fid, mode, m), fids-cons(root, fs, ofh), fd ∈ ofh ⊦ fid ∈ fs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ fids-cons(root, fs, ofh) ↔ (∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids)
Proof step 4:
ofh[fd] = mkfh(fid, mode, m), fd ∈ ofh, ∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids ⊦ fid ∈ fs

Rule argument

spec: posix-base name: fids-cons seq: ⊦ fids-cons(root, fs, ofh) ↔ (∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids) subst: [ofh, fs, root] → [ofh, fs, root] flags: <all paths>

Additional information

Used terms: ofh, fs, root