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Proof step 1 from proof fids-in (conclusion)

⊦ fid ∈ tree.fids ↔ (∃ p. p ∈ tree ∧ tree[p] = fnode(fid))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
fid ∈ tree.fids ⊦ ∃ p. p ∈ tree ∧ tree[p] = fnode(fid)
Proof step 31:
∃ p. p ∈ tree ∧ tree[p] = fnode(fid) ⊦ fid ∈ tree.fids

Rule argument

pos: right 1

Additional information