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Proof step 3 from proof fids-path (previous step)

tree[p] = fnode(fid), p ∈ tree ⊦ fid ∈ tree.fids

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ fid ∈ tree.fids ↔ (∃ p. p ∈ tree ∧ tree[p] = fnode(fid))
Proof step 5:
tree[p] = fnode(fid), p ∈ tree ⊦ ∃ p. p ∈ tree ∧ tree[p] = fnode(fid)

Rule argument

spec: <toplevel> name: fids-in seq: ⊦ fid ∈ tree.fids ↔ (∃ p. p ∈ tree ∧ tree[p] = fnode(fid)) subst: [tree, fid] → [tree, fid] flags: <all paths>

Additional information

Used terms: tree, fid