LTS Termination Proof

by AProVE

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
f4423_0_transfer_ArrayAccess f4423_0_transfer_ArrayAccess f4423_0_transfer_ArrayAccess: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f3804_0_put_NULL f3804_0_put_NULL f3804_0_put_NULL: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f3882_0_put_EQ f3882_0_put_EQ f3882_0_put_EQ: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f3248_0_createMap_LE f3248_0_createMap_LE f3248_0_createMap_LE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f4328_0_transfer_GE f4328_0_transfer_GE f4328_0_transfer_GE: x1 = x1x2 = x2x3 = x3x4 = x4x5 = x5x6 = x6x7 = x7x8 = x8x9 = x9x10 = x10x11 = x11x12 = x12x13 = x13
f517_0_createMap_Return f517_0_createMap_Return f517_0_createMap_Return: 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
f3559_0_clear_GE f3559_0_clear_GE f3559_0_clear_GE: 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
and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 4 SCC(s) of the program graph.

2.1 SCC Subproblem 1/4

Here we consider the SCC { f3248_0_createMap_LE }.

2.1.1 Transition Removal

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

f3248_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 2/4

Here we consider the SCC { f3804_0_put_NULL, f3882_0_put_EQ }.

2.2.1 Transition Removal

We remove transitions 7, 9, 10, 11, 12, 13, 8 using the following ranking functions, which are bounded by 0.

f3804_0_put_NULL: 1 + x4
f3882_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 3/4

Here we consider the SCC { f4423_0_transfer_ArrayAccess, f4328_0_transfer_GE }.

2.3.1 Transition Removal

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

f4328_0_transfer_GE: −1 − x4 + 2⋅x8
f4423_0_transfer_ArrayAccess: −1 − x3 + 2⋅x10

2.3.2 Transition Removal

We remove transition 16 using the following ranking functions, which are bounded by 0.

f4328_0_transfer_GE: −1 − x4 + x7 + x8
f4423_0_transfer_ArrayAccess: −2 − x3 + x10 + x13

2.3.3 Transition Removal

We remove transitions 19, 17 using the following ranking functions, which are bounded by 0.

f4423_0_transfer_ArrayAccess: x4
f4328_0_transfer_GE: −1

2.3.4 Trivial Cooperation Program

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

2.4 SCC Subproblem 4/4

Here we consider the SCC { f3559_0_clear_GE }.

2.4.1 Transition Removal

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

f3559_0_clear_GE: x3 + x6

2.4.2 Trivial Cooperation Program

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

Tool configuration

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