LTS Termination Proof

Input

Integer Transition System

Initial Location: l4, l1, l3, l0

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _result_11HAT0 ∧ x2 = _t_16HAT0 ∧ x3 = _t_22HAT0 ∧ x4 = _temp0_14HAT0 ∧ x5 = _x_13HAT0 ∧ x6 = _y_15HAT0 ∧ x7 = _y_21HAT0 ∧ x1 = _result_11HATpost ∧ x2 = _t_16HATpost ∧ x3 = _t_22HATpost ∧ x4 = _temp0_14HATpost ∧ x5 = _x_13HATpost ∧ x6 = _y_15HATpost ∧ x7 = _y_21HATpost ∧ _y_21HAT0 = _y_21HATpost ∧ _y_15HAT0 = _y_15HATpost ∧ _x_13HAT0 = _x_13HATpost ∧ _temp0_14HAT0 = _temp0_14HATpost ∧ _t_22HAT0 = _t_22HATpost ∧ _t_16HAT0 = _t_16HATpost ∧ _result_11HAT0 = _result_11HATpost
l1	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x7 = _x6 ∧ x1 = _x7 ∧ x2 = _x8 ∧ x3 = _x9 ∧ x4 = _x10 ∧ x5 = _x11 ∧ x6 = _x12 ∧ x7 = _x13 ∧ _x6 = _x13 ∧ _x5 = _x12 ∧ _x4 = _x11 ∧ _x3 = _x10 ∧ _x2 = _x9 ∧ _x1 = _x8 ∧ _x7 = _x3 ∧ _x5 ≤ 0 ∧ 1 ≤ _x4
l1	3	l2:	x1 = _x14 ∧ x2 = _x15 ∧ x3 = _x16 ∧ x4 = _x17 ∧ x5 = _x18 ∧ x6 = _x19 ∧ x7 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ x4 = _x24 ∧ x5 = _x25 ∧ x6 = _x26 ∧ x7 = _x27 ∧ _x20 = _x27 ∧ _x19 = _x26 ∧ _x17 = _x24 ∧ _x16 = _x23 ∧ _x15 = _x22 ∧ _x25 ≤ 0 ∧ _x21 = _x17 ∧ _x25 = _x25 ∧ _x18 ≤ 0
l1	4	l3:	x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ x5 = _x32 ∧ x6 = _x33 ∧ x7 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ x6 = _x40 ∧ x7 = _x41 ∧ _x41 = _x41 ∧ _x37 = _x37 ∧ 1 ≤ _x32 ∧ 1 ≤ _x33 ∧ _x42 = _x32 ∧ _x43 = −2 + _x33 ∧ _x44 = 1 + _x42 ∧ _x36 = _x43 ∧ _x39 = −2 + _x44 ∧ _x40 = 1 + _x36 ∧ _x39 ≤ −1 + _x37 ∧ −1 + _x37 ≤ _x39 ∧ _x40 ≤ 1 + _x36 ∧ 1 + _x36 ≤ _x40 ∧ _x36 ≤ −2 + _x41 ∧ −2 + _x41 ≤ _x36 ∧ 1 ≤ _x41 ∧ 1 ≤ _x37 ∧ _x28 = _x35 ∧ _x31 = _x38
l3	5	l1:	x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ x4 = _x48 ∧ x5 = _x49 ∧ x6 = _x50 ∧ x7 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ x5 = _x56 ∧ x6 = _x57 ∧ x7 = _x58 ∧ _x51 = _x58 ∧ _x50 = _x57 ∧ _x49 = _x56 ∧ _x48 = _x55 ∧ _x47 = _x54 ∧ _x46 = _x53 ∧ _x45 = _x52
l4	6	l0:	x1 = _x59 ∧ x2 = _x60 ∧ x3 = _x61 ∧ x4 = _x62 ∧ x5 = _x63 ∧ x6 = _x64 ∧ x7 = _x65 ∧ x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x4 = _x69 ∧ x5 = _x70 ∧ x6 = _x71 ∧ x7 = _x72 ∧ _x65 = _x72 ∧ _x64 = _x71 ∧ _x63 = _x70 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x59 = _x66

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7

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

l1:	−1 + x5 + x6
l3:	−1 + x5 + x6

2.1.2 Transition Removal

We remove transition 5 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 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)