(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
cond(tt, x, y) → f(s(x), s(y))
f(x, y) → cond(lt(x, y), x, y)
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
COND(tt, x, y) → F(s(x), s(y))
F(x, y) → COND(lt(x, y), x, y)
F(x, y) → LT(x, y)
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
cond(tt, x, y) → f(s(x), s(y))
f(x, y) → cond(lt(x, y), x, y)
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LT(s(x), s(y)) → LT(x, y)
The TRS R consists of the following rules:
cond(tt, x, y) → f(s(x), s(y))
f(x, y) → cond(lt(x, y), x, y)
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- LT(s(x), s(y)) → LT(x, y)
The graph contains the following edges 1 > 1, 2 > 2
(7) YES
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, y) → COND(lt(x, y), x, y)
COND(tt, x, y) → F(s(x), s(y))
The TRS R consists of the following rules:
cond(tt, x, y) → f(s(x), s(y))
f(x, y) → cond(lt(x, y), x, y)
lt(0, y) → tt
lt(s(x), s(y)) → lt(x, y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) NonLoopProof (EQUIVALENT transformation)
By Theorem 8 [NONLOOP] we deduce infiniteness of the QDP.
We apply the theorem with m = 1, b = 1,
σ' = [ ], and μ' = [ ] on the rule
COND(tt, zr0, zr1)[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0] → COND(tt, s(zr0), s(zr1))[zr0 / s(zr0), zr1 / s(zr1)]n[zr0 / 0]
This rule is correct for the QDP as the following derivation shows:
intermediate steps: Equivalent (Remove Unused) - Instantiate mu - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
COND(tt, x1, x0)[x1 / s(x1), x0 / s(x0), zt2 / s(zt2), zt3 / s(zt3)]n[x1 / 0, x0 / y0, zt2 / 0, zt3 / y0] → COND(tt, s(zt2), s(zt3))[x1 / s(x1), x0 / s(x0), zt2 / s(zt2), zt3 / s(zt3)]n[x1 / 0, x0 / y0, zt2 / 0, zt3 / y0]
by Narrowing at position: []
intermediate steps: Equivalent (Add Unused) - Instantiate mu - Equivalent (Add Unused) - Instantiate Sigma - Instantiation
COND(tt, x, y)[ ]n[ ] → F(s(x), s(y))[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming) - Equivalent (Remove Unused)
F(s(zl2), s(zl3))[zl2 / s(zl2), zl3 / s(zl3), zr2 / s(zr2), zr3 / s(zr3)]n[zl2 / 0, zl3 / x0, zr2 / 0, zr3 / x0] → COND(tt, s(zr2), s(zr3))[zl2 / s(zl2), zl3 / s(zl3), zr2 / s(zr2), zr3 / s(zr3)]n[zl2 / 0, zl3 / x0, zr2 / 0, zr3 / x0]
by Narrowing at position: [0]
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
F(s(zs2), s(zs3))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0] → COND(lt(y1, y0), s(zs2), s(zs3))[zs2 / s(zs2), zs3 / s(zs3)]n[zs2 / y1, zs3 / y0]
by Narrowing at position: [0]
intermediate steps: Instantiate mu - Instantiate Sigma - Instantiation - Instantiation - Instantiation
F(x, y)[ ]n[ ] → COND(lt(x, y), x, y)[ ]n[ ]
by OriginalRule from TRS P
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Equivalent (Domain Renaming) - Instantiation - Equivalent (Domain Renaming)
lt(s(x), s(y))[x / s(x), y / s(y)]n[ ] → lt(x, y)[ ]n[ ]
by PatternCreation I
lt(s(x), s(y))[ ]n[ ] → lt(x, y)[ ]n[ ]
by OriginalRule from TRS R
intermediate steps: Equivalent (Add Unused) - Equivalent (Add Unused) - Instantiation - Instantiation
lt(0, y)[ ]n[ ] → tt[ ]n[ ]
by OriginalRule from TRS R
(10) NO