LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l1, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_5000HAT0 ∧ x2 = _x_13HAT0 ∧ x3 = _x_27HAT0 ∧ x4 = _x_32HAT0 ∧ x5 = _y_16HAT0 ∧ x6 = _y_28HAT0 ∧ x7 = _y_33HAT0 ∧ x1 = ___const_5000HATpost ∧ x2 = _x_13HATpost ∧ x3 = _x_27HATpost ∧ x4 = _x_32HATpost ∧ x5 = _y_16HATpost ∧ x6 = _y_28HATpost ∧ x7 = _y_33HATpost ∧ _y_33HAT0 = _y_33HATpost ∧ _y_28HAT0 = _y_28HATpost ∧ _x_32HAT0 = _x_32HATpost ∧ _x_27HAT0 = _x_27HATpost ∧ ___const_5000HAT0 = ___const_5000HATpost ∧ _y_16HATpost = _y_16HATpost ∧ _x_13HATpost = _x_13HATpost
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 ∧ _x3 = _x10 ∧ _x2 = _x9 ∧ _x1 = _x8 ∧ _x = _x7 ∧ 1 ≤ _x1 ∧ 1 ≤ _x11 ∧ _x11 = _x ∧ 1 ≤ _x1
l3	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 ∧ _x17 = _x24 ∧ _x14 = _x21 ∧ 1 ≤ _x26 ∧ 1 ≤ _x23 ∧ −1 + _x26 ≤ _x25 ∧ _x25 ≤ −1 + _x26 ∧ −1 + _x23 ≤ _x22 ∧ _x22 ≤ −1 + _x23 ∧ _x25 = −1 + _x18 ∧ _x22 = −1 + _x15 ∧ 1 ≤ _x18 ∧ _x26 = _x26 ∧ _x23 = _x23
l2	4	l1:	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 ∧ _x34 = _x41 ∧ _x33 = _x40 ∧ _x32 = _x39 ∧ _x31 = _x38 ∧ _x30 = _x37 ∧ _x29 = _x36 ∧ _x28 = _x35 ∧ _x32 ≤ 0 ∧ _x32 ≤ 0
l2	5	l4:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ x7 = _x48 ∧ x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x7 = _x55 ∧ _x47 = _x54 ∧ _x44 = _x51 ∧ _x42 = _x49 ∧ 1 ≤ _x55 ∧ −1 + _x55 ≤ _x53 ∧ _x53 ≤ −1 + _x55 ∧ −1 + _x52 ≤ _x50 ∧ _x50 ≤ −1 + _x52 ∧ _x53 = −1 + _x46 ∧ _x50 = −1 + _x43 ∧ 1 ≤ _x46 ∧ _x55 = _x55 ∧ _x52 = _x52
l4	6	l2:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x5 = _x60 ∧ x6 = _x61 ∧ x7 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ x4 = _x66 ∧ x5 = _x67 ∧ x6 = _x68 ∧ x7 = _x69 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x59 = _x66 ∧ _x58 = _x65 ∧ _x57 = _x64 ∧ _x56 = _x63
l5	7	l0:	x1 = _x70 ∧ x2 = _x71 ∧ x3 = _x72 ∧ x4 = _x73 ∧ x5 = _x74 ∧ x6 = _x75 ∧ x7 = _x76 ∧ x1 = _x77 ∧ x2 = _x78 ∧ x3 = _x79 ∧ x4 = _x80 ∧ x5 = _x81 ∧ x6 = _x82 ∧ x7 = _x83 ∧ _x76 = _x83 ∧ _x75 = _x82 ∧ _x74 = _x81 ∧ _x73 = _x80 ∧ _x72 = _x79 ∧ _x71 = _x78 ∧ _x70 = _x77

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
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
l2	l2	l2:	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 { l4, l1, l2 }.

2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

l1:	−1 + x1 + x2
l2:	−1 + x1 + x2 − x5
l4:	−1 + x1 + x2 − x5

2.1.2 Transition Removal

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

l2:	0
l1:	−1
l4:	0

2.1.3 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

l4:	2⋅x5 + 1
l2:	2⋅x5

2.1.4 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

l4:	0
l2:	−1

2.1.5 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 % (8 real / 0 unknown / 0 assumptions / 8 total proof steps)