We consider the problem of reducing the (semi)total domination number of graph by one by contracting edges. It is known that this can always be done with at most three edge contractions and that deciding whether one edge contraction suffices is an NP-hard problem. We show that for every fixed k ∈ {2, 3}, deciding whether exactly k edge contractions are necessary is NP-hard and further provide for k = 2 complete complexity dichotomies on monogenic graph classes.Next, consider a clause c ∈ C and let x, y, z ∈ X be the variables contained in c. Then |D ∩ {u c , v c , w c , q x c , q y c , q z c }| ≤ 2 for D ∩ {u c , v c , w c , q x c , q y c , q z c } would otherwise contain a friendly triple; and we conclude by Claim 24 that, in fact, equality holds. Now suppose that D ∩{f x c , f y c , f z c } = ∅, say f x c ∈ D without loss of generality. Then q x c / ∈ D: indeed, D would otherwise contain the friendly triple f x c , q x c , t, where t ∈ D ∩ {u c , v c , w c , q y c , q z c }, a contradiction to Theorem 1.7. But by assumption, P cx,F (2) is not a private neighbour of f x c and the vertex in D ∩ {P c x,F (1), P c x,F (2)} is witnessed by P cx,F (4), which implies thatc , q y c , q z c }| ≥ 3, a contradiction to the above. Thus,∈ D without loss of generality, then surely one of u x c and v x c does not belong to D (D would otherwise contain the friendly triple t x c , u x c , v x c ), say u x c / ∈ D without loss of generality. Since by assumption, P cx,T (1) is not a private neighbour of t x c and the vertex in D ∩ {P c x,T (1), P c x,T (2)} is witnessed by P cx,T (4), the set. By repeating this argument if necessary, we obtain a minimum SD sety, z}} contains a friendly triple, a contradiction to Theorem 1.7. Thus, |D ∩ V (G T c )| ≤ 2 and we conclude by Claim 24(ii) that, in fact, equality holds. Therefore, |D ∩ V (G c )| = 4 and thus, γ t2 (G) = |D| = 14|X| + 4|C|. c and t y c should be dominated, necessarilycontains an edge by assumption, it must be that u a c , u b c ∈ D or v a c , v b c ∈ D for two variables a, b ∈ {x, y, z}. In the first case, since D ∩ {v x c , v y c , v z c } = ∅ by Claim 24(ii), necessarily D ∩ {t x c , t y c , t z c } = ∅; and in the case, since D ∩ {u x c , u y c , u z c } = ∅ by Claim 24(ii), necessarily D ∩ {t x c , t y c , t z c } = ∅. Thus, in both cases, we conclude that D contains no edge satisfying (aT) or (bT); but then D contains only one edge, a contradiction. Thus D)) contains no edge. Since by assumption {t x c } ∪ V (P c x,T ) and {t y c } ∪ V (P c y,T ) contain no edge as well, it follows that D ∩ V (G T c ) must contain a P 3 (D would otherwise contain only one edge). However, since25, which implies that no edge of an ST-configuration satisfies (aT) or (bT). Let us next show that no edge of an STconfiguration in D satisfies (cT). Suppose that D ∩ V (G T c ) contains an edge uv. Then by ??(ii), {u, v} = {u ℓ c , v ℓ c } for some ℓ ∈ {x, y, z}, say x without loss of generality. If P c x,T (1) / ∈ D then d({u, v}, D \ {u, v} ≥ 3 and so, uv cannot be part of an ST-configurati...