From KLs obtained as a product of three pretzel tangles, those of the types [1] [1] [1], [0] [1] [1], [1] [0] [0], [1] [¥] [1], [0] [0] [1], [1] [0] [0], [0] [1] ¥], [1] [0] [¥], [¥] [0] [1], [¥] [1] [0], [1] [¥] [¥], [¥] [1] [¥], [¥] [¥] [1], [0] [0] [¥], [¥] [0] [0], [¥] [¥] [¥], are knots. From the symmetry reasons, it is sufficient to consider the knots of the type [1] [1] [1], [0] [1] [1], [1] [¥] [1], [0] [0] [1], [0] [1] ¥], [1] [0] [¥], [1] [¥] [¥], [¥] [1] [¥], [0] [0] [¥], [¥] [¥] [¥].

For example, a knot of the type [1] [1] [1] will be non-invertible iff the pretzel tangle in the middle is not mirror-symmetrical, or if it is mirror-symmetrical and the first pretzel tangle is different from the last one and from its reverse. As this example shows, it is still possible to find general conditions for non-invertibility of certain classes and types of knots, but the conditions will be more complicated then before and dependent from the types of knots considered, and not only from classes they belong.

A KL of the form (r₁,r₂) (2k-1) (r₃,r₄), where r_i (i=1,¼,4) are R-tangles, and k=1,2,¼ is a knot iff its type is [1] [1] [1], [0] [1] [1], [1] [1] [0], [0] [1] [¥], [¥] [1] [0], [¥] [1] [¥]. From the symmetry reasons, it is sufficient to consider knots of the type [1] [1] [1], [0] [1] [1], [0] [1] [¥], [¥] [1] [¥]. A knot of the type [1] [1] [1], [¥] [1] [¥] is non-invertible iff its pretzel parts (r₁,r₂) and (r₃,r₄) are different, and invertible otherwise. A knot of the type [0] [1] [¥] is non-invertible iff r₁ ¹ r₂, and invertible otherwise. A knot of the type [0] [1] [1] is always non-invertible.

A KL of the form (r₁,r₂) (2k) (r₃,r₄), where r_i (i=1,¼,4) are R-tangles, and k=1,2,¼is a knot iff its type is [0] [0] [1], [1] [0] [0], [1] [0] [¥], [¥] [0] [1], [0] [0] [¥], [¥] [0] [0]. From the symmetry reasons, it is sufficient to consider knots of the type [0] [0] [1], [1] [0] [¥], [0] [0] [¥]. A knot of that form is non-invertible iff it does not contain a [¥] tangle, and invertible otherwise. Hence, all non-invertible knots are of the type [0] [0] [1].

An arborescent KL of the form (r₁,r₂),r₃,(r₄,r₅), where r_i (i=1,¼,5) are R-tangles, is a knot iff its (pretzel) type is [0,¥,0], [0,¥,1], [1,¥,1], [0,1,0], [1,1,1], [¥,1,0], [¥,1,1], [1,0,¥], [0,0,¥], [0,0,1]. A knot of the type [0,¥,0], [0,¥,1], [1,¥,1] (with ¥ in the middle) is non-invertible iff the tangles (r₁,r₂), (r₄,r₅) are different. A knot of the type [0,1,0], [1,1,1], [¥,1,0], [¥,1,1] (with 1 in the middle) is non-invertible iff the tangle (r₁,r₂) is different from (r₄,r₅) and from its reverse (r₅,r₄). A knot of the type [1,0,¥], [0,0,¥], [0,0,1] (with 0 in the middle) is always non-invertible.

The next class we consider are KLs of the form p₁,p₂,¼,p_i, where p_k (k=3,4,¼) are pretzel tangles of the form (r₁,r₂,¼,r_j) (j=2,3,¼). The result is a knot iff among p_k tangles there is an odd number of the tangles of the pretzel type [1] and an arbitrary number of the tangles of the pretzel type [¥], or if there is exactly one tangle of the pretzel type [0]. The simplest case of such knots are those of the form: (r₁,r₂),(r₃,r₄),(r₅,r₆). They are non-invertible iff they do not contain equal pretzel tangles, i.e., iff their t-graph is not mirror-symmetrical. For example, knot (5,2),(3,2),(2 1,2 1) is non-invertible, and (3,2),(3,2),(2 1,2 1) is invertible. Non-invertible knots (3,4),(3,2),(2 1,2 1), (5,2),(3,2),(2 1,2 1) belonging to the same family are obtained by breaking symmetry of its corresponding t-diagram. In general, a knot of the form p₁,p₂,¼,p_i is non-invertible iff its t-diagram is not mirror-symmetrical. For example, knots (2 1,3),(2 1,3),(3,2),(3,2 1), (2 1,3),(2 1,3),(2 1,3),(3,2 1),(2 1,5), (2 1,3),(2 1,3),(2 1,3),(2 1,5),(4 1,3) are non-invertible, and (2 1,3),(3,2),(2 1,3),(3,2 1) is invertible.