/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.pl /export/starexec/sandbox/output/output_files


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YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.pl
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Left Termination of the query pattern

e(g)

w.r.t. the given Prolog program could successfully be proven:

(0) Prolog
(1) PrologToDTProblemTransformerProof [SOUND, 0 ms]
(2) TRIPLES
(3) TPisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


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(0)
Obligation:
Clauses:

e(X) :- ','(=(Y, .(e, .(o, Y))), c(Y, X)).
c(.(e, X1), 0).
c(.(X2, X), s(Y)) :- c(X, Y).


Query: e(g)
----------------------------------------

(1) PrologToDTProblemTransformerProof (SOUND)
Built DT problem from termination graph DT10.

{
    "root": 59,
    "program": {
        "directives": [],
        "clauses": [
            [
                "(e X)",
                "(',' (= Y (. (e) (. (o) Y))) (c Y X))"
            ],
            [
                "(c (. (e) X1) (0))",
                null
            ],
            [
                "(c (. X2 X) (s Y))",
                "(c X Y)"
            ]
        ]
    },
    "graph": {
        "nodes": {
            "59": {
                "goal": [{
                    "clause": -1,
                    "scope": -1,
                    "term": "(e T1)"
                }],
                "kb": {
                    "nonunifying": [],
                    "intvars": {},
                    "arithmetic": {
                        "type": "PlainIntegerRelationState",
                        "relations": []
                    },
                    "ground": ["T1"],
                    "free": [],
                    "exprvars": []
                }
            },
            "60": {
                "goal": [{
                    "clause": 0,
                    "scope": 1,
                    "term": "(e T1)"
                }],
                "kb": {
                    "nonunifying": [],
                    "intvars": {},
                    "arithmetic": {
                        "type": "PlainIntegerRelationState",
                        "relations": []
                    },
                    "ground": ["T1"],
                    "free": [],
                    "exprvars": []
                }
            },
            "61": {
                "goal": [{
                    "clause": -1,
                    "scope": -1,
                    "term": "(',' (= X5 (. (e) (. (o) X5))) (c X5 T3))"
                }],
                "kb": {
                    "nonunifying": [],
                    "intvars": {},
                    "arithmetic": {
                        "type": "PlainIntegerRelationState",
                        "relations": []
                    },
                    "ground": ["T3"],
                    "free": ["X5"],
                    "exprvars": []
                }
            },
            "type": "Nodes",
            "62": {
                "goal": [],
                "kb": {
                    "nonunifying": [],
                    "intvars": {},
                    "arithmetic": {
                        "type": "PlainIntegerRelationState",
                        "relations": []
                    },
                    "ground": [],
                    "free": [],
                    "exprvars": []
                }
            }
        },
        "edges": [
            {
                "from": 59,
                "to": 60,
                "label": "CASE"
            },
            {
                "from": 60,
                "to": 61,
                "label": "ONLY EVAL with clause\ne(X4) :- ','(=(X5, .(e, .(o, X5))), c(X5, X4)).\nand substitutionT1 -> T3,\nX4 -> T3"
            },
            {
                "from": 61,
                "to": 62,
                "label": "UNIFY-FAIL\nbecause of non-unification"
            }
        ],
        "type": "Graph"
    }
}

----------------------------------------

(2)
Obligation:
Triples:


Clauses:


Afs:


----------------------------------------

(3) TPisEmptyProof (EQUIVALENT)
There are no more dependency triples. Hence, the dependency triple problem trivially terminates.
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(4)
YES