Completion Proof

by kbcv

Input

For the following equations E

a(1,1) = 1
a(1,0) = 0
a(0,1) = 0
a(0,0) = 0
o(1,1) = 1
o(1,0) = 1
o(0,1) = 1
o(0,0) = 0
n(1) = 0
n(0) = 1
n(n(g)) = g
a(g,1) = g
a(1,g) = g
o(g,0) = g
o(0,g) = g
a(g,0) = 0
a(0,g) = 0
o(g,1) = 1
o(1,g) = 1
a(g,n(g)) = 0
a(n(g),g) = 0
o(g,n(g)) = 1
o(n(g),g) = 1
a(g,g) = g
o(g,g) = g

and the following TRS R

0 → n(1)
n(n(g)) → g
a(g,1) → g
a(1,g) → g
o(g,1) → 1
o(1,g) → 1
o(g,n(g)) → 1
o(n(g),g) → 1
a(g,g) → g
o(g,g) → g
o(n(1),g) → g
o(g,n(1)) → g
a(g,n(1)) → n(1)
a(n(1),g) → n(1)
a(n(g),g) → n(1)
a(x,n(x)) → n(1)

it is proven that E is equivalent to R, and R is convergent.

Proof

1 Completion Proof

We have to prove termination and local confluence of R, and equivalence of R and E.

1.1 Rule Removal

Using the recursive path order with the following precedence and status
prec(1) = 0 stat(1) = lex
prec(n) = 1 stat(n) = lex
prec(0) = 2 stat(0) = lex
prec(a) = 2 stat(a) = lex
all rules could be removed.

1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.

1.2 Local Confluence Proof

All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.

1.3 Equivalence Proof of R and E

R can be simulated by E as follows. All rules could be derived from the equations

That E can be simulated by R can be shown by just computing normal forms of each equation in E.

Tool configuration

kbcv