In NMTs, an empty proof is defined as a proof that satisfies two conditions: (1) it contains an empty range i.e., proof.start == proof.end, and (2) its nodes field is empty. However, currently, we do not check whether proof.leafHash == 0. This can lead to situations where a proof can be treated as both an absence proof and an empty proof, depending on the order of operations (depends on which type is checked first). Specifically, a proof that satisfies conditions (1) and (2) but has a non-empty leafHash will pass both IsEmptyProof() and IsOfAbsence() checks.
The following test demonstrates this inconsistency:
// create an NMT with 8 sequentially namespaced leaves, numbered from 1 to 8.
tree := exampleNMT(1, 1, 2, 3, 4, 5, 6, 7, 8)
root, err := tree.Root()
assert.NoError(t, err)
dummyValue := createByteSlice(tree.treeHasher.Size(), 0x01)
// this proof is neither a well-formed absence proof nor a well-formed empty proof
proof := Proof{start: 1, end: 1, leafHash: dummyValue, nodes: [][]byte{}}
assert.True(t, proof.IsOfAbsence()) // this check passes while proof is not a well-formed absence proof, this result is misleading
assert.True(t, proof.IsEmptyProof()) // this check passes while proof is not a well-formed empty proof, this result is misleading
// the proof is not a proper empty proof, yet it can be used as a valid empty proof
isVerified := proof.VerifyNamespace(tree.treeHasher.baseHasher, []byte{10}, [][]byte{}, root)
assert.True(t, isVerified)
Problem
In NMTs, an empty
proofis defined as a proof that satisfies two conditions: (1) it contains an empty range i.e.,proof.start == proof.end, and (2) itsnodesfield is empty. However, currently, we do not check whetherproof.leafHash == 0. This can lead to situations where a proof can be treated as both an absence proof and an empty proof, depending on the order of operations (depends on which type is checked first). Specifically, a proof that satisfies conditions (1) and (2) but has a non-emptyleafHashwill pass bothIsEmptyProof()andIsOfAbsence()checks.The following test demonstrates this inconsistency: