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Proof step 3 from proof fids-cons-reduce (previous step)

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))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ fids-cons(root, fs, ofh) ↔ (∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids)
Proof step 5:
∀ fid. fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids ⊦ fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids))

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] → [ofh0, fs0, tree0] flags: <all paths>

Additional information

Used terms: ofh0, fs0, tree0