pub proof fn lemma_prefix_of_common_sequence(
source1: Seq<nat>,
source2: Seq<nat>,
child: Seq<nat>,
)Expand description
requires
source1.is_prefix_of(child),source2.is_prefix_of(child),ensuressource1 == source2 || source1.len() < source2.len() && source1.is_prefix_of(source2)
|| source2.len() < source1.len() && source2.is_prefix_of(source1),If source1 and source2 are prefixes of child, then either source1 is equal to source2 or
one of them is a prefix of the other.