Certification Problem

Input (TPDB TRS_Standard/SK90/4.47)

The rewrite relation of the following TRS is considered.

f(g(i(a,b,b'),c),d) → if(e,f(.(b,c),d'),f(.(b',c),d')) (1)
f(g(h(a,b),c),d) → if(e,f(.(b,g(h(a,b),c)),d),f(c,d')) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1, x2)] = 2 · x1 + 2 · x2
[g(x1, x2)] = 2 · x1 + 1 · x2
[i(x1, x2, x3)] = 1 + 1 · x1 + 2 · x2 + 1 · x3
[a] = 0
[b] = 0
[b'] = 0
[c] = 0
[d] = 0
[if(x1, x2, x3)] = 2 · x1 + 2 · x2 + 2 · x3
[e] = 0
[.(x1, x2)] = 1 · x1 + 2 · x2
[d'] = 0
[h(x1, x2)] = 2 · x1 + 1 · x2
all of the following rules can be deleted.
f(g(i(a,b,b'),c),d) → if(e,f(.(b,c),d'),f(.(b',c),d')) (1)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(g(h(a,b),c),d) → f#(.(b,g(h(a,b),c)),d) (3)
f#(g(h(a,b),c),d) → f#(c,d') (4)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.