LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
f3797_0_createMap_LE f3797_0_createMap_LE f3797_0_createMap_LE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f4358_0_put_InvokeMethod f4358_0_put_InvokeMethod f4358_0_put_InvokeMethod: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f3875_0_random_ArrayAccess f3875_0_random_ArrayAccess f3875_0_random_ArrayAccess: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f4529_0_put_EQ f4529_0_put_EQ f4529_0_put_EQ: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f5087_0_transfer_ArrayAccess f5087_0_transfer_ArrayAccess f5087_0_transfer_ArrayAccess: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f1_0_main_Load f1_0_main_Load f1_0_main_Load: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f617_0_createMap_Return f617_0_createMap_Return f617_0_createMap_Return: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
__init __init __init: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f4448_0_put_NULL f4448_0_put_NULL f4448_0_put_NULL: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f4991_0_transfer_GE f4991_0_transfer_GE f4991_0_transfer_GE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

2.1 SCC Subproblem 1/3

Here we consider the SCC { f3797_0_createMap_LE }.

2.1.1 Transition Removal

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

f3797_0_createMap_LE: x2

2.1.2 Trivial Cooperation Program

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

2.2 SCC Subproblem 2/3

Here we consider the SCC { f4529_0_put_EQ, f4448_0_put_NULL }.

2.2.1 Transition Removal

We remove transitions 8, 10, 11, 12, 13, 14, 9 using the following ranking functions, which are bounded by 0.

f4448_0_put_NULL: 1 + x4
f4529_0_put_EQ: x3

2.2.2 Trivial Cooperation Program

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

2.3 SCC Subproblem 3/3

Here we consider the SCC { f5087_0_transfer_ArrayAccess, f4991_0_transfer_GE }.

2.3.1 Transition Removal

We remove transition 19 using the following ranking functions, which are bounded by 0.

f4991_0_transfer_GE: −1 − x4 + 2⋅x8
f5087_0_transfer_ArrayAccess: −1 − x3 + 2⋅x10

2.3.2 Transition Removal

We remove transition 17 using the following ranking functions, which are bounded by 0.

f4991_0_transfer_GE: −1 − x4 + x7 + x8
f5087_0_transfer_ArrayAccess: −2 − x3 + x10 + x13

2.3.3 Transition Removal

We remove transitions 20, 18 using the following ranking functions, which are bounded by 0.

f5087_0_transfer_ArrayAccess: 4⋅x4
f4991_0_transfer_GE: x1

2.3.4 Trivial Cooperation Program

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

Tool configuration

AProVE