LTS Termination Proof

Input

Integer Transition System

Initial Location: l4, l1, l3, l0

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _iEXCL14HAT0 ∧ x2 = _iEXCL22HAT0 ∧ x3 = _resultEXCL12HAT0 ∧ x4 = _temp0EXCL15HAT0 ∧ x1 = _iEXCL14HATpost ∧ x2 = _iEXCL22HATpost ∧ x3 = _resultEXCL12HATpost ∧ x4 = _temp0EXCL15HATpost ∧ _iEXCL14HAT1 = 0 ∧ 0 ≤ _iEXCL14HAT1 ∧ _iEXCL14HAT1 ≤ 0 ∧ 0 ≤ _iEXCL14HAT1 ∧ _iEXCL14HAT1 ≤ 0 ∧ 1 + _iEXCL14HAT1 ≤ 10 ∧ _iEXCL14HAT2 = 1 + _iEXCL14HAT1 ∧ 1 ≤ _iEXCL14HAT2 ∧ _iEXCL14HAT2 ≤ 1 ∧ 1 ≤ _iEXCL14HAT2 ∧ _iEXCL14HAT2 ≤ 1 ∧ 1 + _iEXCL14HAT2 ≤ 10 ∧ _iEXCL14HATpost = 1 + _iEXCL14HAT2 ∧ 2 ≤ _iEXCL14HATpost ∧ _iEXCL14HATpost ≤ 2 ∧ _iEXCL22HAT0 = _iEXCL22HATpost ∧ _resultEXCL12HAT0 = _resultEXCL12HATpost ∧ _temp0EXCL15HAT0 = _temp0EXCL15HATpost
l1	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x1 = _x5 ∧ _x = _x4 ∧ _x6 = _x3 ∧ 10 ≤ _x
l1	3	l3:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ 1 + _x9 ≤ 10 ∧ 1 + _x9 ≤ _x12 ∧ _x12 ≤ 1 + _x9 ∧ _x12 = 1 + _x8 ∧ 1 + _x8 ≤ 10
l3	4	l1:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x16 = _x20
l4	5	l0:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x24 = _x28

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4

and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l1, l3 }.

2.1.1 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

l1:	18 − x1 − x2
l3:	18 − x1 − x2

2.1.2 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

l3:	0
l1:	−1

2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE

version: AProVE Commit ID: unknown
strategy: Statistics for single proof: 100.00 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)