YES

Problem 1:

(VAR vu95NonEmpty x z)
(RULES
c -> t(k)
c -> t(l)
f(x) -> z | s(x) ->* z
g(x,x) -> h(x,x)
s(a) -> c
s(b) -> c
)

Problem 1:

Well-founded Relation Processor:
-> Rules:
 c -> t(k)
 c -> t(l)
 f(x) -> z | s(x) ->* z
 g(x,x) -> h(x,x)
 s(a) -> c
 s(b) -> c
->AGES Output:

Model Results

System:
mod InTheory is
sort S .
sort Bool .


op _->*_ : S S -> Bool [m = 2] .
op _->_ : S S -> Bool [m = 2] .
op c :  -> S .
op f : S -> S .
op g : S S -> S .
op s : S -> S .
op a :  -> S .
op b :  -> S .
op fSNonEmpty :  -> S .
op h : S S -> S .
op k :  -> S .
op l :  -> S .
op t : S -> S .
op sqsupset : S S -> Bool [wellfounded m = 1] .

endm


Property:
x ->R* x
x ->R y /\ y ->R* z => x ->R* z
x1 ->R y1 => f(x1) ->R f(y1)
x1 ->R y1 => g(x1,x2) ->R g(y1,x2)
x2 ->R y2 => g(x1,x2) ->R g(x1,y2)
x1 ->R y1 => s(x1) ->R s(y1)
x1 ->R y1 => h(x1,x2) ->R h(y1,x2)
x2 ->R y2 => h(x1,x2) ->R h(x1,y2)
x1 ->R y1 => t(x1) ->R t(y1)
c ->R t(k)
c ->R t(l)
s(x) ->R* z => f(x) ->R z
g(x,x) ->R h(x,x)
s(a) ->R c
s(b) ->R c
x ->R y => sqsupset(x,y)

Results:


Domains:
S: |N

Function Interpretations:
|[a]| = 2
|[b]| = 2
|[c]| = 1
|[f(x_1_1:S)]| = 2 + x_1_1:S
|[fSNonEmpty]| = 1
|[g(x_1_1:S,x_2_1:S)]| = 3 + 3.x_1_1:S + x_2_1:S
|[h(x_1_1:S,x_2_1:S)]| = x_1_1:S + 3.x_2_1:S
|[k]| = 0
|[l]| = 0
|[s(x_1_1:S)]| = x_1_1:S
|[t(x_1_1:S)]| = x_1_1:S

Predicate Interpretations:
 x_1_1:S ->* x_2_1:S <=> ((1 + x_1_1:S + x_2_1:S >= 0) /\ (1 + x_1_1:S >= x_2_1:S))
 x_1_1:S -> x_2_1:S <=> ((x_1_1:S >= 1 + x_2_1:S) /\ (1 + x_1_1:S >= x_2_1:S))
sqsupset(x_1_1:S,x_2_1:S) <=> (3.x_1_1:S >= 1 + 3.x_2_1:S)

The problem is finite.
2.90user 0.12system 0:03.31elapsed 91%CPU (0avgtext+0avgdata 79804maxresident)k
25296inputs+152outputs (113major+22152minor)pagefaults 0swaps