L = {w1w2| there are two words, x,y such that : xw1 is in L1, w2y is in L2} is regular if L1 and L2 are regular languages.
Lsuff = { w1 | xw1 ∈ L1 }
Lpref = { w2 | w2y ∈ L2 }
And,
L = LsuffLpref
We can easily proof by construction Finite Automata for L
.
Suppose Finite Automata(FA) for L1 is M1 and FA for L2 is M2.
[SOLUTION]
Non-Deterministic Finite Automata(NFA) for L can be drawn by introducing NULL-transition (^-edge) form every state in M1 to every state in M2. then NFA can be converted into DFA.
e.g.
L1 = {ab ,ac} and L2 = {12, 13}
L = {ab, ac, 12, 13, a12, a2, ab12, ab2, a13, a3, ab13, ab3, ............}
Note: w1 and w2 can be NULL
M1 =is consist of Q = {q0,q1,qf} with edges:
q0 ---a----->q1,
q1 ---b/c--->qf
Similarly :
M2 =is consist of Q = {p0,p1,pf} with edges:
p0 ---1----->p1,
p1 ---2/3--->pf
Now, NFA for L called M will be consist of Q = {q0,q1,qf, p0,p1,pf} Where Final state of M is pf and edges are:
q0 ---a----->q1,
q1 ---b/c--->qf,
p0 ---1----->p1,
p1 ---2/3--->pf,
q0 ----^----> p0,
q1 ----^----> p0,
qf ----^----> p0,
q0 ----^----> p1,
q1 ----^----> p1,
qf ----^----> p1,
q0 ----^----> pf,
q1 ----^----> pf,
qf ----^----> pf
^ means NULL-Transition.
Now, A NFA can easily convert into DFA.(I leave it for you)
[ANSWER]
DFA for L is possible hence L is Regular Language.
I will highly encourage you to draw DFA/NFA figures, then concept will be clear.