YES

We show the termination of the TRS R:

  eq(|0|(),|0|()) -> true()
  eq(|0|(),s(x)) -> false()
  eq(s(x),|0|()) -> false()
  eq(s(x),s(y)) -> eq(x,y)
  or(true(),y) -> true()
  or(false(),y) -> y
  union(empty(),h) -> h
  union(edge(x,y,i),h) -> edge(x,y,union(i,h))
  reach(x,y,empty(),h) -> false()
  reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
  if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
  if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
  if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
  if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: eq#(s(x),s(y)) -> eq#(x,y)
p2: union#(edge(x,y,i),h) -> union#(i,h)
p3: reach#(x,y,edge(u,v,i),h) -> if_reach_1#(eq(x,u),x,y,edge(u,v,i),h)
p4: reach#(x,y,edge(u,v,i),h) -> eq#(x,u)
p5: if_reach_1#(true(),x,y,edge(u,v,i),h) -> if_reach_2#(eq(y,v),x,y,edge(u,v,i),h)
p6: if_reach_1#(true(),x,y,edge(u,v,i),h) -> eq#(y,v)
p7: if_reach_2#(false(),x,y,edge(u,v,i),h) -> or#(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
p8: if_reach_2#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,h)
p9: if_reach_2#(false(),x,y,edge(u,v,i),h) -> reach#(v,y,union(i,h),empty())
p10: if_reach_2#(false(),x,y,edge(u,v,i),h) -> union#(i,h)
p11: if_reach_1#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,edge(u,v,h))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The estimated dependency graph contains the following SCCs:

  {p3, p5, p8, p9, p11}
  {p1}
  {p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: if_reach_2#(false(),x,y,edge(u,v,i),h) -> reach#(v,y,union(i,h),empty())
p2: reach#(x,y,edge(u,v,i),h) -> if_reach_1#(eq(x,u),x,y,edge(u,v,i),h)
p3: if_reach_1#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,edge(u,v,h))
p4: if_reach_1#(true(),x,y,edge(u,v,i),h) -> if_reach_2#(eq(y,v),x,y,edge(u,v,i),h)
p5: if_reach_2#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,h)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The set of usable rules consists of

  r1, r2, r3, r4, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_reach_2#_A(x1,x2,x3,x4,x5) = x1 + x3 + x4 + ((0,1),(0,0)) x5
        false_A() = (3,1)
        edge_A(x1,x2,x3) = ((0,0),(1,0)) x2 + ((0,1),(0,1)) x3 + (1,0)
        reach#_A(x1,x2,x3,x4) = x2 + ((0,1),(0,0)) x3 + ((0,1),(0,0)) x4 + (3,0)
        union_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (1,0)
        empty_A() = (0,0)
        if_reach_1#_A(x1,x2,x3,x4,x5) = x3 + ((0,1),(0,0)) x4 + ((0,1),(0,0)) x5 + (3,0)
        eq_A(x1,x2) = ((0,0),(1,1)) x1 + ((1,0),(1,0)) x2 + (1,1)
        true_A() = (1,8)
        |0|_A() = (3,1)
        s_A(x1) = ((1,1),(1,1)) x1 + (3,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if_reach_2#_A(x1,x2,x3,x4,x5) = ((0,1),(0,0)) x1 + ((0,1),(1,0)) x3 + ((0,1),(0,1)) x4
        false_A() = (4,4)
        edge_A(x1,x2,x3) = (2,2)
        reach#_A(x1,x2,x3,x4) = ((0,1),(0,0)) x2 + (5,3)
        union_A(x1,x2) = (1,1)
        empty_A() = (1,1)
        if_reach_1#_A(x1,x2,x3,x4,x5) = ((0,1),(0,0)) x3 + (5,3)
        eq_A(x1,x2) = ((1,1),(1,1)) x2 + (1,1)
        true_A() = (4,4)
        |0|_A() = (1,1)
        s_A(x1) = ((1,1),(0,0)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p4, p5

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: reach#(x,y,edge(u,v,i),h) -> if_reach_1#(eq(x,u),x,y,edge(u,v,i),h)
p2: if_reach_1#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,edge(u,v,h))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: reach#(x,y,edge(u,v,i),h) -> if_reach_1#(eq(x,u),x,y,edge(u,v,i),h)
p2: if_reach_1#(false(),x,y,edge(u,v,i),h) -> reach#(x,y,i,edge(u,v,h))

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        reach#_A(x1,x2,x3,x4) = x2 + ((0,1),(0,0)) x3 + x4 + (1,0)
        edge_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((0,1),(1,1)) x2 + ((1,0),(1,1)) x3 + (1,3)
        if_reach_1#_A(x1,x2,x3,x4,x5) = x3 + ((0,1),(0,0)) x4 + x5
        eq_A(x1,x2) = x2 + (1,1)
        false_A() = (1,1)
        |0|_A() = (1,1)
        true_A() = (0,0)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        reach#_A(x1,x2,x3,x4) = ((0,0),(1,1)) x2 + ((1,1),(0,1)) x4
        edge_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((1,0),(1,1)) x3 + (1,2)
        if_reach_1#_A(x1,x2,x3,x4,x5) = x3 + (1,1)
        eq_A(x1,x2) = x2 + (1,1)
        false_A() = (3,2)
        |0|_A() = (1,1)
        true_A() = (3,2)
        s_A(x1) = (3,1)
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: eq#(s(x),s(y)) -> eq#(x,y)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        eq#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(1,1)) x2
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        eq#_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,0),(1,1)) x2
        s_A(x1) = ((0,1),(0,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: union#(edge(x,y,i),h) -> union#(i,h)

and R consists of:

r1: eq(|0|(),|0|()) -> true()
r2: eq(|0|(),s(x)) -> false()
r3: eq(s(x),|0|()) -> false()
r4: eq(s(x),s(y)) -> eq(x,y)
r5: or(true(),y) -> true()
r6: or(false(),y) -> y
r7: union(empty(),h) -> h
r8: union(edge(x,y,i),h) -> edge(x,y,union(i,h))
r9: reach(x,y,empty(),h) -> false()
r10: reach(x,y,edge(u,v,i),h) -> if_reach_1(eq(x,u),x,y,edge(u,v,i),h)
r11: if_reach_1(true(),x,y,edge(u,v,i),h) -> if_reach_2(eq(y,v),x,y,edge(u,v,i),h)
r12: if_reach_2(true(),x,y,edge(u,v,i),h) -> true()
r13: if_reach_2(false(),x,y,edge(u,v,i),h) -> or(reach(x,y,i,h),reach(v,y,union(i,h),empty()))
r14: if_reach_1(false(),x,y,edge(u,v,i),h) -> reach(x,y,i,edge(u,v,h))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        union#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        edge_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        union#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        edge_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
    

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

We remove them from the problem.  Then no dependency pair remains.