**Proof of Theorem 1.3**

Let *G* be a graph satisfying the condition of Theorem 1.3. Let *C* be a longest cycle of *G* and assign an orientation to it. Suppose *G* is not Hamiltonian. Then $V(G)\backslash V(C)\ne \overline{)0}$. Let *R* be a component of *G* − *C*, and *A* = {*υ*_{1}, *υ*_{2}, … , *υ*_{k}} be the set of neighbors of *R* on *C*. Since *G* is 2-connected, there exists a (*υ*_{i}, *υ*_{j})-path *P* = *υ*_{i}u_{1} … *u*_{r} υ_{j} with all internal vertices in *R*, and *υ*_{i}, *υ*_{j} ∈ *A*. Choose *P* such that:

*There is no* o-*cycle C*′ *in G such that V*(*C*) ⊂ *V*(*C*′).

*Proof*. Otherwise, *C*′ is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′). By Lemma 2.2, there exists a cycle containing all vertices in *C*′ and longer than *C*, contradicting the choice of *C*. □

By Lemma 2.3 (*e*), without loss of generality, assume that *υ*^{−}_{i}*υ*^{+}_{i} ∈ *E*(*G*).

*r* = 1, *that is*, *V*(*P*) = {*υ*_{j}, *u*_{1}, *υ*_{j}}.

*Proof*. Suppose *r* ≥ 2. Consider *H* = *G*[*V*(*P*) ∪ *V*(*C*[*υ*_{i}, *υ*_{j}])] − {*υ*_{j}}. Since *υ*^{−}_{i}*υ*^{+}_{i} ∈ *E*(*G*), *υ*_{i}*υ*^{−}_{j} ∉ *E*(*G*) by Lemma 2.3 (*c*). Thus *d*_{H}(*υ*^{−}_{j}, *υ*_{i}) ≥ 2. By the choice condition of *P* and Lemma 2.3 (*a*), we have *d*_{P}(*υ*_{i}, *u*_{r}) ≥ 2 and *d*_{H}(*υ*^{−}_{j}, *u*_{r}) = *d*_{H}(*υ*^{−}_{j}, *υ*_{i}) + *d*_{H}(*υ*_{j}, *u*_{r}) ≥ 4, which yields a contradiction to the fact *G* is almost distance-hereditary and *d*_{G}(*υ*^{−}_{j}, *u*_{r}) = 2. Hence *V*(*P*) = {*υ*_{i}, *u*_{1}, *υ*_{j}}. □

|*V*(*C*[*υ*_{i}, *υ*_{j}])| ≥ 5.

*Proof*. Suppose |*V*(*C*[*υ*_{i}, *v*_{j}])| = 4 or |*V*(*C*[*υ*_{i}, *υ*_{j}])| = 3. This means *C*[*υ*_{i}, *υ*_{j}] = *υ*_{i}*υ*^{+}_{i}*υ*^{−}_{i}*υ*_{j} or *C*[*υ*_{i}, *υ*_{j}] = *υ*_{i}*υ*^{−}_{i}*υ*_{j}. Let *C*′ = *υ*_{i}*u*_{1}*υ*_{j}*υ*^{−}_{j}*υ*^{+}_{j}*C*[*υ*^{+}_{j}, *υ*^{−}_{i}]*υ*^{−}_{i}*υ*^{+}_{i}*υ*_{i} or *C*′ = *υ*_{i}*u*_{1}*υ*_{j}*υ*^{−}_{j}*υ*^{+}_{j}*C*[*υ*^{+}_{j}, *υ*_{i}]. Then *C*′ is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′) by Lemma 2.3 (*d*), contradicting Claim 1. □

Recall that *υ*^{–}_{i} *υ*^{+}_{i} ∈ *E*(*G*). Let *H* = *G*[{*u*_{1}, *υ*^{–}_{i}} ∪ *V*(*C*[*υ*_{i}, *υ*_{j}])] – {*υ*_{i}}. Since *d*_{G}(*υ*^{–}_{i}, *u*_{1}) = 2 and *G* is almost distance-hereditary, *d*_{H}(*υ*^{–}_{i}, *u*_{1}) ≤ 3. By Lemma 2.3 (*c*) and (*d*), we have *υ*^{–}_{i} *υ*_{j} ∉ *E*(*G*). By the choice of *P*, *u*_{1}*υ* ∉ *E*(*G*), where *υ* ∈ *C*[*υ*^{+}_{i}, *υ*^{–}_{j}]. It follows that *d*_{H}(*υ*^{–}_{i}, *u*_{1}) = 3 and *d*_{H}(*υ*^{–}_{i}, *υ*_{j}) = 2. By Lemma 2.3 (*b*), (*c*) and (*d*), *υ*^{–}_{i} *υ*^{–}_{j} ∉ *E*(*G*) and *υ*^{+}_{i} *υ*_{j} ∉ *E*(*G*). Thus there exists a vertex *w* ∈ *C*(*υ*^{+}_{i}, *υ*^{–}_{j}) such that *υ*^{–}_{i} *w* ∈ *E*(*G*) and *wυ*_{j} ∈ *E*(*G*). Note that *w* is well-defined.

*wυ*^{+}_{j} ∉ *Ẽ*(*G*).

*Proof*. Suppose *wυ*^{+}_{j} ∉ *Ẽ*(*G*). By Lemma 2.4 (*a*), we obtain *υ*^{–}_{i} *w*^{+} ∉ *Ẽ*(*G*). Since *υ*^{–}_{i} *w* ∈ *E*(*G*), we have *υ*_{j}w^{+} ∉ *Ẽ*(*G*) by Lemma 2.4 (*b*) and by symmetry. Note that *υ*_{j} *υ*^{–}_{i} ∉ *Ẽ*(*G*) by Lemma 2.3 (*c*). Thus {*w*, *w*^{+}, *υ*_{j}, *υ*^{–}_{i}} induces an *o*-light claw in *G*, a contradiction.

Next we will show that {*υ*_{j}, *u*_{1}, *w*, *υ*^{+}_{j}} induces an *o*-light claw and get a contradiction. Before proving this fact, the following claim is needed.

*Proof*. First we will show that *w*^{–}*υ*^{–}_{i} ∉ *Ẽ*(*G*). Since *υ*^{–}_{j} *υ*^{+}_{j} ∈ *Ẽ*(*G*) and *υ*_{j} w ∈ *E*(*G*), we have *w*^{–} *υ*^{–}_{i} ∉ *Ẽ*(*G*) by Lemma 2.4 (*b*) and symmetry.

Next we will show that *w*^{−} *υ*_{j} ∈ *Ẽ*(*G*). Suppose not. Consider the subgraph induced by {*w*, *w*^{−}, *υ*_{j}, *υ*^{–}_{i}}. Note that *υ*_{j} *υ*^{–}_{i} ∉ *Ẽ*(*G*) by Lemma 2.3 (*c*) and *w*^{–} *υ*^{–}_{i} ∉ *Ẽ*(*G*) by the analysis above. Then {*w*, *w*^{−}, *υ*_{j}, *υ*^{–}_{i}} induces an *o*-light claw, a contradiction.

Now we will show that *u*_{1} *w* ∉ *Ẽ*(*G*), since otherwise, *C*′ = *u*_{1}*wC*[*w*, *υ*^{–}_{j}]*υ*^{–}_{j} *υ*^{+}_{j}*C*[*υ*^{+}_{j}, *w*^{–}]*w*^{–} *υ*_{j}u_{1} is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), contradicting Claim 1. □

By Claims 4, 5 and Lemma 2.3 (*a*), {*υ*_{j}, *u*_{1}, *w*, *υ*^{+}_{j}} induces an *o*-light claw, contradicting the fact *G* is claw-heavy. The proof of Theorem 1.3 is complete. □

**Proof of Theorem 1.4.**

Let *G* be a graph satisfying the condition of Theorem 1.4. By Lemma 2.1, there exists a heavy cycle in *G*. Now choose a longest heavy cycle *C* of *G* and assign an orientation to it. Suppose *G* is not Hamiltonian. Then $V(G)\backslash V(C)\ne \overline{)0}$. Let *R* be a component of *G* − *C* and *A* = {*w*_{1}, *w*_{2}, … , *w*_{k}} be the set of neighbors of *R* on *C*. Since *G* is 3-connected, for any vertex *u* of *R*, there exists a (*u*, *C*; 3)-fan *F* such that *F* = (*u*; *Q*_{1}, *Q*_{2}, *Q*_{3}), where *Q*_{1} = *ux*_{1} … *x*_{r1} *w*_{i}, *Q*_{2} = *uy*_{1} … *y*_{r2} *w*_{j} and *Q*_{3} = *uz*_{1} … *z*_{r3} *w*_{k}, and *w*_{i}, *w*_{j}, *w*_{k} are in the order of the orientation of *C*.

By the choice of *C*, all internal vertices of *F* are not heavy. By Lemma 2.3 (*b*), there is at most one heavy vertex in *N*^{+}_{C}(*R*) and at most one heavy vertex in *N*^{–}_{C}(*R*). Without loss of generality, assume that *w*^{–}_{i}, *w*^{+}_{i} are light. Hence *w*^{–}_{i} *w*^{+}_{i} ∈ *E*(*G*), otherwise {*w*_{i}, *w*^{–}_{i}, *w*^{+}_{i}, *x*_{r1}} induces a light claw, contradicting *G* is 1-heavy.

*There exists a* (*u*, *C*; 3)-*fan F such that V*(*F*) = {*u*, *w*_{i}, *w*_{j}, *w*_{k}}.

*Proof*. Now we choose the fan *F* in such a way that:

(1)

*Q*_{1} = *uw*_{i};

(2)

|*V*(*C*[*w*_{i}, *w*_{j}])| is as small as possible subject to (1);

(3)

|*V*(*Q*_{2})| is as small as possible subject to (1) and (2);

(4)

|*V*(*C*[*w*_{k}, *w*_{i}])| is as small as possible subject to (1), (2) and (3);

(5)

|*V*(*Q*_{3})| is as small as possible subject to (1), (2), (3) and (4).

Since *G* is 3-connected, for any neighbor of *C* in *R*, say *u* (with *uw*_{i} ∈ *E*(*G*), where *w*_{i} ∈ *V*(*C*)), there are three disjoint paths from *u* to *C*. Obviously, we can choose one such path as *uw*_{i}. Thus (1) is well-defined, and furthermore, the choice condition of *F* is well-defined.

*Proof*. Suppose $V({Q}_{2})\backslash \{u,{w}_{j}\}\ne \overline{)0}$. Without loss of generality, set *y* = *y*_{r2}. Let *H* = *G*[*V*(*Q*_{1}) ∪ *V*(*Q*_{2}) ∪ *V*(*C*[*w*_{i}, *w*_{j}])] − {*w*_{j}}. Note that *w*^{–}_{i} *w*^{+}_{i} ∈ *E*(*G*). By Lemma 2.3 (*c*), it is easy to see that *w*_{i} w^{–}_{j} ∉ *E*(*G*), so *d*_{H}(*w*^{–}_{j}, *w*_{i}) ≥ 2. In the meantime, the choice condition (2) implies that $N(V({Q}_{2})\backslash \{{w}_{j}\})\cap V(C({w}_{j},{w}_{j}))=\overline{)0}$. This means that *d*_{H} (*w*^{–}_{j}, *y*) = *d*_{H} (*w*^{–}_{j}, *w*_{i}) + *d*_{H} (*w*_{i}, *y*) ≥ 2 + *d*_{H} (*w*_{i}, *y*). Since *G* is almost distance-hereditary and *d*_{G} (*w*^{–}_{j}, *y*) = 2, we have *d*_{H} (*w*^{–}_{j}, *y*) = 3 and *yw*_{i} ∈ *E*(*G*). Let *F*′ = (*y*; *Q*′_{1}, *Q*′_{2}, *Q*′_{3}) such that *Q*′_{1} = *yw*_{i}, *Q*′_{2} = *yw*_{j} and *Q*′_{3} = *Q*_{2}[*y*, *u*]*Q*_{3}[*u*, *w*_{k}]. Then *F*′ is a (*y*, *C*;3)-fan satisfying (1), (2) and |*V*(*Q*′_{2})| = 2, contradicting the choice condition (3), a contradiction. □

*Proof*. Suppose $V({Q}_{\text{3}})\backslash \{u,{w}_{k}\}\ne \overline{)0}$. Without loss of generality, set *z* = *z*_{r3}.

If *zw*_{i} ∉ *E*(*G*), then set *H* = *G*[*V* (*Q*_{1}) ∪ *V* (*Q*_{3}) ∪ *V*(*C*[*w*_{k}, *w*_{i}])] − {*w*_{k}}. Since *w*^{–}_{i} *w*^{+}_{i} ∈ *E*(*G*), we obtain *w*^{+}_{k} *w*_{i} ∉ *E*(*G*) by Lemma 2.3 (*c*). This means *d*_{H} (*w*^{+}_{k}, *w*_{i}) ≥ 2. By the choice condition (4), we have $N(V({Q}_{3})\backslash \{{w}_{k}\})\cap V(C({w}_{k},{w}_{i}))=\overline{)0}$, and hence *d*_{H} (*w*^{+}_{k}, *z*) = *d*_{H} (*w*^{+}_{k}, *w*_{i}) + *d*_{H} (*w*_{i}, *z*). Since *zw*_{i} ∉ *E*(*G*), *d*_{H} (*w*_{i}, *z*) ≥ 2 and we get *d*_{H} (*w*^{+}_{k}, *z*) ≥ 4. It yields a contradiction to the fact *G* is almost distance-hereditary and *d*_{G} (*w*^{+}_{k}, *z*) = 2.

If *zw*_{i} ∈ *E*(*G*), then set *H* = *G*[*V*(*C* [*w*_{i}, *w*_{j}]) ∪ *V*(*Q*_{3} [*u*, *z*])] − {*w*_{i}}. Note that ${N}_{C}(z)\cap V(C({w}_{i},{w}_{j}])=\overline{)0}$ (by the choice conditions (2), (5)) and ${N}_{C}(V({Q}_{3})\backslash \{z,{w}_{k}\})\cap V(C({w}_{i},{w}_{j}))=\overline{)0}$ (by the choice condition (2)). Since *d*_{G}(*w*^{+}_{i}, *z*) = 2 and *G* is almost distance-hereditary, *d*_{H}(*w*^{+}_{i}, *z*) ≤ 3. But if *w*^{+}_{i}, *w*_{j} ∉ *E*(*G*), then the distance from *z* to *w*^{+}_{i} in *H* is at least 4, where in such a shortest path, the path *Q*_{3} [*z*, *u*] contributes at least 1, the path *Q*_{2} [*u*, *w*_{j}] contributes 1, a contradiction. Thus we have *w*^{+}_{i} *w*_{j} ∈ *E*(*G*), and hence *w*^{–}_{j} *w*^{+}_{j} ∉ *Ẽ*(*G*) by Lemma 2.3 (*c*). Consider the subgraph induced by {*w*_{j}, *w*^{+}_{i}, *w*^{+}_{j}, *u*}. Since *G* is 1-heavy and *w*^{+}_{i}, *u* are light, *w*^{+}_{j} is heavy. Now let *H* = *G*[{*w*^{–}_{i}} ∪ *V*(*C*[*w*_{i}, *w*_{j}]) ∪ *V*(*Q*_{3}[*u*, *z*])] – {*w*_{i}}. Similarly, since *d*_{G}(*w*^{–}_{i}, *z*) = 2, we have *d*_{H}(*w*^{–}_{i}, *z*) = 3, and *w*^{–}_{i}, *w*_{j} ∈ *E*(*G*). Consider the subgraph induced by {*w*_{j}, *w*^{–}_{i}, *w*^{–}_{j}, *u*}. Similarly, we can see *w*^{–}_{j} is heavy, and hence *w*^{–}_{j} *w*^{+}_{j} ∈ *Ẽ*(*G*), a contradiction. Thus *V*(*Q*_{3}) = {*u*, *w*_{k}}. □

By Claims 1.1 and 1.2, the proof of Claim 1 is complete. □

By Claim 1, there exists a (*u*, *C*; 3)-fan *F* such that *V*(*F*)\*V*(*C*) = {*u*}. Suppose that *N*_{C} (*u*) = {*υ*_{1}, *υ*_{2}, …, *υ*_{r}} (*r* ≥ 3) and *υ*_{1}, *υ*_{2}, …, *υ*_{r} are in the order of the orientation of *C*. In the following, all the subscripts of *υ* are taken modulo *r*, and *υ*_{0} = *υ*_{r}.

By Lemma 2.3 (*b*), there is at most one heavy vertex in *N*^{+}_{C}(*u*) and at most one heavy vertex in *N*^{–}_{C} (*u*). Since *r* ≥ 3, we know that there exists *υ*_{j} ∈ *N*_{C} (*u*), such that *υ*^{–}_{j}, *υ*^{+}_{j} are light, and hence *υ*^{–}_{j} *υ*^{+}_{j} ∈ *E*(*G*) by the fact *G* is 1-heavy. Without loss of generality, assume that *υ*^{–}_{1} *υ*^{+}_{1} ∈ *E*(*G*) and *υ*^{–}_{1}, *υ*^{+}_{1} are light. By Lemma 2.6 (*b*), there exists a vertex *l*_{1} ∈ *C*[*υ*^{+}_{1}, *υ*^{–}_{2}) such that *υ*^{–}_{2}*l*_{1} ∈ *E*(*G*) and *l*_{1}*υ*_{1} ∈ *E*(*G*), and there exists a vertex *s*_{1} ∈ *C*(*υ*^{+}_{0}, *υ*^{–}_{1}] such that *υ*^{+}_{0} *s*_{1} ∈ *E*(*G*) and *s*_{1} *v*_{1} ∈ *E*(*G*).

We divide the proof into two cases.

*υ*^{–}_{2}*υ*^{+}_{2} ∉ *E*(*G*) *and* *υ*^{–}_{0} *μ*^{+}_{0} ∉ *E*(*G*).

Both {*υ*, *υ*^{–}_{2}, *υ*^{+}_{2}, *u*} and {*υ*, *υ*^{–}_{0}, *υ*^{+}_{0}, *u*} induce claws. By Lemma 2.3 (*b*) and the fact *G* is 1-heavy, *υ*^{–}_{2} and *υ*^{+}_{0} are heavy or *υ*^{+}_{2} and *υ*^{–}_{0} are heavy.

*υ*^{–}_{1}*l*_{1} ∈ *E*(*G*) *and* *l*_{1} *υ*_{2} ∈ *E*(*G*).

*Proof*. Suppose *υ*^{–}_{1}*l*_{1} ∉ *E*(*G*). Note that *uv*^{–}_{1} ∉ *Ẽ*(*G*) by Lemma 2.3 (*a*). By Lemma 2.5 (*a*), *l*_{1} is light. Now {*υ*_{1}, *l*_{1}, *u*, *υ*^{–}_{1}} induces a light claw, a contradiction.

Suppose *l*_{1} *v*_{2} ∉ *E*(*G*). Let *H* = *G*[{*υ*^{–}_{1}, *l*_{1}, *υ*^{–}_{2}, *υ*_{2}, *u*}]. By Lemma 2.3, we get *uv*^{–}_{2} ∉ *E*(*G*), *uv*^{–}_{1} ∉ *E*(*G*) and *υ*^{–}_{1} *υ*^{–}_{2} ∉ *E*(*G*). Note that *υ*_{2}*υ*^{–}_{1} ∉ *E*(*G*) by Lemma 2.5 (a). Now *G*[{*υ*^{–}_{1}, *l*_{1}, *υ*^{–}_{2}, *υ*_{2}, *u*}] is an induced path of length 4 in *G*, contradicting Lemma 2.6 (*a*).

Now we consider the following two subcases.

*υ*^{–}_{2}, *υ*^{+}_{0} *are heavy vertices.*

By Lemma 2.3 (*a*), *uυ*^{+}_{2} ∉ *Ẽ*(*G*). By Lemma 2.5 (*a*), we have *υ*_{1}*l*^{–}_{1} ∈ *E*(*G*). Note that *l*′ ≔ *l*^{–}_{1} ∈ *N*(*υ*_{1}) and *υ*^{–}_{1}*υ*^{+}_{1} ∈ *E*(*G*). By Lemma 2.4 (*b*), *l*′^{+}*υ*^{+}_{2} = *l*_{1}*υ*^{+}_{2} ∉ *Ẽ*(*G*). By Lemma 2.5 (*a*) and Lemma 2.3 (*b*), *l*_{1} and *υ*^{+}_{2} are light. Now {*υ*_{2}, *l*_{1}, *u*, *υ*^{+}_{2}} induces a light claw, a contradiction.

*υ*^{+}_{2}, *υ*^{–}_{0} *are heavy vertices.*

Consider the subgraph induced by {*υ*^{+}_{0}, *s*_{1}, *l*_{1}, *υ*_{2}, *υ*}. It is easily to check that *υ*^{+}_{0}*s*_{1} ∈ *E*(*G*), *l*_{1}*υ*_{2} ∈ *E*(*G*) (by Claim 2) and *υ*_{2}*u* ∈ *E*(*G*). By Lemma 2.3 (*a*), *υ*^{+}_{0}*u* ∉ *E*(*G*). By Lemma 2.5 (*a*) and Lemma 2.4 (*b*), we know that *υ*_{1}*l*^{–}_{1} ∈ *E*(*G*) and *υ*^{+}_{0}*l*_{1} ∉ *E*(*G*). Now we obtain *s*_{1}*l*_{1} ∉ *E*(*G*) or *υ*^{+}_{0}*υ*_{2} ∈ *E*(*G*) or *s*_{1}*υ*_{2} ∈ *E*(*G*) (Otherwise, *G*[{*υ*^{+}_{0}, *s*_{1}, *l*_{1}, *υ*_{2}, *u*}] is an induced path of length 4, contradicting Lemma 2.6 (*a*)).

Suppose *s*_{1}*l*_{1} ∉ *E*(*G*). By Lemma 2.5 (*a*), *l*_{1}, *s*_{1} are light. Now {*υ*_{1}, *l*_{1}, *s*_{1}, *u*} induces a light claw, contradicting *G* is 1-heavy.

Suppose *υ*^{+}_{0}*υ*_{2} ∈ *E*(*G*). Consider the subgraph induced by {*υ*_{2}, *υ*^{–}_{2}, *υ*^{+}_{0}, *u*}. By Lemma 2.3 (*a*), we have *uυ*^{–}_{2} ∉ *E*(*G*) and *uυ*^{+}_{0} ∉ *E*(*G*). Since *υ*^{–}_{2}, *υ*^{+}_{0}, *u* are light and *G* is 1-heavy, *υ*^{+}_{0}*υ*^{–}_{2} ∈ *E*(*G*). By Lemma 2.5 (*a*) and Claim 2, *υ*_{1}*l*^{–}_{1} ∈ *E*(*G*) and *l*_{1}*υ*_{2} ∈ *E*(*G*). Now ${C}^{\prime}={\upsilon}_{1}{l}_{1}^{-}\overleftarrow{C}[{l}_{1}^{-},{\upsilon}_{1}^{+}]{\upsilon}_{1}^{+}{\upsilon}_{1}^{-}\overleftarrow{C}[{\upsilon}_{1}^{-},{\upsilon}_{0}^{+}]{\upsilon}_{0}^{+}{\upsilon}_{2}^{-}\overleftarrow{C}[{\upsilon}_{2}^{-},{l}_{1}]{l}_{1}{\upsilon}_{2}C[{\upsilon}_{2},{\upsilon}_{0}]{\upsilon}_{0}u{\upsilon}_{1}$ is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), a contradiction.

Suppose *s*_{1}*υ*_{2} ∈ *E*(*G*). Consider the subgraph induced by {*υ*_{2}, *υ*^{–}_{2}, *s*_{1}, *u*}. By Lemma 2.5 (*a*) and Lemma 2.3 (*b*), *s*_{1} and *υ*^{–}_{2} are light. Since *G* is 1-heavy, *s*_{1}*υ*^{–}_{2} ∈ *E*(*G*). Now ${C}^{\prime}={\upsilon}_{1}u{\upsilon}_{2}C[{\upsilon}_{2},{s}_{1}]{s}_{1}{\upsilon}_{2}^{-}\overleftarrow{C}[{\upsilon}_{2}^{-},{\upsilon}_{1}^{+}]{\upsilon}_{1}^{+}{s}_{1}^{+}C[{s}_{1}^{+},{\upsilon}_{1}]$ is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), a contradiction. (First, we can prove *s*_{1}*υ*^{+}_{1} ∈ *E*(*G*). Otherwise, {*υ*_{1}, *s*_{1}, *υ*^{+}_{1}, *u*} induces a light claw, a contradiction. Note that *υ*^{+}_{0}*υ*^{+}_{1} ∉ *E*(*G*) and *υ*^{+}_{0}*s*^{+}_{1} ∉ *E*(*G*) by Lemma 2.3 (*b*) and Lemma 2.4 (*b*). Then we obtain *υ*^{+}_{1}*s*^{+}_{1} ∈ *E*(*G*) since otherwise {*s*_{1}, *υ*^{+}_{0}, *υ*^{+}_{1}, *s*^{+}_{1}} induces a light claw, a contradiction.)

*υ*^{–}_{2} *υ*^{+}_{2} ∈ *E*(*G*) *or* *υ*^{–}_{0} *υ*^{+}_{0} ∈ *E*(*G*).

Without loss of generality (by symmetry), assume that *υ*^{–}_{2} *υ*^{+}_{2} ∈ *E*(*G*).

*υ*^{–}_{0} *υ*^{+}_{0} ∉ *E*(*G*).

By Lemma 2.3 (*a*), *uυ*^{–}_{0} ∉ *E*(*G*) and *uυ*^{+}_{0} ∉ *E*(*G*). Now {*υ*_{0}, *υ*^{–}_{0}, *υ*^{+}_{0}, *u*} induces a claw. Since *G* is 1-heavy and *u* is light, *υ*^{–}_{0} is heavy or *υ*^{+}_{0} is heavy.

Suppose *υ*^{–}_{0} is heavy. By Lemma 2.3 (*b*), (*c*) and Lemma 2.4 (*b*), *υ*^{–}_{2}, *υ*_{1} and *l*^{–}_{1} are light. By Lemma 2.6 (*d*), {*l*_{1}, *l*^{–}_{1}, *υ*_{1}, *υ*^{–}_{2}} induces a light claw, a contradiction.

Suppose *υ*^{+}_{0} is heavy. By Lemma 2.3 (*b*), (*c*) and Lemma 2.4 (*b*), we can see *υ*^{+}_{2}, *υ*_{1} and *l*^{+}_{1} are light. By Lemma 2.6 (*d*), {*l*_{1}, *l*^{+}_{1}, *υ*_{1}, *υ*^{+}_{2}} induces a light claw, a contradiction. (Note that *υ*^{–}_{1}*υ*^{+}_{1} ∈ *E*(*G*) and *υ*^{–}_{2}*υ*^{+}_{2} ∈ *E*(*G*). By Lemma 6 (*c*), *l*_{1}*υ*^{+}_{2} ∈ *E*(*G*).

*υ*^{–}_{0}*υ*^{+}_{0} ∈ *E*(*G*).

By Lemma 6 (*b*), there exists a vertex *s*_{2} ∈ *C*(*υ*^{+}_{1}, *υ*^{–}_{2}] such that *υ*^{+}_{1}*s*_{2} ∈ *E*(*G*) and *s*_{2}*υ*_{2} ∈ *E*(*G*).

(*i*) *υ*_{1} *is heavy*, (*ii*) *l*^{–}_{0}, *l*^{+}_{0}, *s*^{–}_{2}, *s*^{+}_{2} *are light.*

*Proof.* Recall that the definition of *l*_{0} occurred in the condition of Lemma 5 before. Let *l*_{0} ∈ *C*[*υ*^{+}_{0}, *υ*^{–}_{1}) such that *υ*^{–}_{0}*l*_{0} ∈ *E*(*G*) and *l*_{0}*υ*_{0} ∈ *E*(*G*).

(i) By Lemma 2.6 (*d*), each of {*l*_{0}, *l*^{–}_{0}, *υ*_{0}, *υ*^{–}_{1}} and {*l*_{1}, *l*^{–}_{1}, *υ*_{1}, *υ*^{–}_{2}} induces a claw. Since *G* is 1-heavy, at least one vertex of {*l*^{–}_{1}, *υ*_{1}, *υ*^{–}_{2}} is heavy.

Suppose *υ*^{–}_{2} is heavy. By Lemma 2.3 (*b*), (*c*) and Lemma 2.4 (*b*), *υ*^{–}_{1}, *υ*_{0} and *l*^{–}_{0} are light. Now {*l*_{0}, *l*^{–}_{0}, *υ*_{0}, *υ*^{–}_{1}} induces a light claw, contradicting *G* is 1-heavy.

Suppose *l*^{–}_{1} is heavy. By Lemma 2.6 (*c*), *υ*^{+}_{2}*l*_{1} ∈ *E*(*G*). By Lemma 2.4 (*a*) and (*b*), *υ*^{–}_{1}*l*^{–}_{1} ∉ *Ẽ*(*G*) and *υ*_{0}*l*^{–}_{1} ∉ *Ẽ*(*G*). This implies that *υ*_{0}, *υ*^{–}_{1} are light. At the same time, we can prove that *l*_{0}^{–} is light. (Otherwise, $C\prime ={\upsilon}_{0}u{\upsilon}_{2}\overleftarrow{C}[{\upsilon}_{2},{l}_{1}]{l}_{1}\text{\hspace{0.17em}}{\upsilon}_{2}^{-}C[{\upsilon}_{2}^{+},{\upsilon}_{0}^{-}]{\upsilon}_{0}^{-}{\upsilon}_{0}^{+}C[{\upsilon}_{0}^{+},{l}_{0}^{-}]{l}_{0}^{-}{l}_{1}^{-}\overleftarrow{C}[{l}_{1}^{-},{l}_{0}]l{\upsilon}_{0}$ is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), a contradiction.) Now {*l*_{0}, *l*^{–}_{0}, *υ*_{0}, *υ*^{–}_{1}} induces a light claw, contradicting *G* is 1-heavy.

Note that {*l*_{1}, *l*^{–}_{1}, *υ*_{1}, *υ*^{–}_{2}} induces a claw and *υ*^{–}_{2}, *l*^{–}_{1} are light. Since *G* is 1-heavy, *υ*_{1} is heavy.

(*ii*) Note that *υ*_{1} is heavy. If *l*^{+}_{0} is heavy, then *C*′ = *υ*_{0}*uυ*_{1}*l*^{+}_{0}*C*[*l*^{+}_{0}, *υ*^{–}_{1}]*υ*^{–}_{1}*υ*^{+}_{1}*C*[*υ*^{+}_{1}, *υ*^{–}_{0}]*υ*^{–}_{0}*υ*^{+}_{0}*C*[*υ*^{+}_{0}, *l*_{0}]*l*_{0}*υ*_{0} is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), a contradiction. If *l*^{–}_{0} is heavy, then *C*′ = *υ*_{0}*l*_{0}*C*[*l*_{0}, *υ*^{–}_{1}]*υ*^{–}_{1}*υ*^{+}_{1}*C*[*υ*^{+}_{1}, *υ*^{–}_{0}]*υ*^{–}_{0}*υ*^{+}_{0}*C*[*υ*^{+}_{0}, *l*^{–}_{0}]*l*^{–}_{0}*υ*_{1}*uυ*_{0} is an *o*-cycle such that *V*(*C*) ⊂ *V*(*C*′), a contradiction. Similarly, by symmetry, we can prove that *s*^{–}_{2}, *s*^{+}_{2} are light. □

*υ*^{–}_{1}*l*^{+}_{0} ∈ *E*(*G*) *and* *υ*^{–}_{1}*s*^{+}_{2} ∈ *E*(*G*).

*Proof.* Suppose *υ*^{–}_{1}*l*^{+}_{0} ∉ *E*(*G*). By Lemma 2.4 (*b*) and (*c*), *υ*^{–}_{1}*l*^{–}_{0} ∉ *Ẽ*(*G*) and *l*^{–}_{0}*l*^{+}_{0} ∉ *Ẽ*(*G*). By Claim 3, *l*^{+}_{0}, *l*^{–}_{0} are light. Now {*l*_{0}, *l*^{–}_{0}, *l*^{+}_{0}, *υ*^{–}_{1}} induces a light claw, a contradiction.

By Lemma 2.6 (*c*) and by symmetry, we obtain *υ*^{–}_{1}*s*_{2} ∈ *E*(*G*). Suppose *υ*^{–}_{1}*s*^{+}_{2} ∉ *E*(*G*). By Lemma 2.4 (*b*) and (*c*), *υ*^{–}_{1}*s*^{–}_{2} ∉ *Ẽ*(*G*) and *s*^{–}_{2}*s*^{+}_{2} ∉ *Ẽ*(*G*). By Claim 3, *s*^{+}_{2}, *s*^{–}_{2} are light. Now {*s*_{2}, *s*^{–}_{2}, *s*^{+}_{2}, *υ*^{–}_{1}} induces a light claw, a contradiction. □

By Claim 3, Lemma 2.5 (*b*) and Claim 4, *l*^{+}_{0}, *s*^{+}_{2} are light, $\{{\upsilon}_{1}^{+}{l}_{0}^{+},{\upsilon}_{1}^{+}{s}_{2}^{+},{l}_{0}^{+}{s}_{2}^{+}\}\cap E(G)=\overline{)0}$ and {*υ*^{–}_{1}*l*^{+}_{0}, *υ*^{–}_{1}*s*^{+}_{2}} ⊂ *E*(*G*). It is proved that {*υ*^{–}_{1}, *υ*^{+}_{1}, *l*^{+}_{0}, *s*^{+}_{0}} induces a light claw, a contradiction.

The proof of Theorem 1.4 is complete. □

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