MAYBE *** Computating Strongly Quasi-Reducible Parts *** TRS: [ eq(?x,?x) -> true, eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true, neq(?x,?y) -> not(eq(?x,?y)), neq(?x,s(?x)) -> true ] Constructors: {0,s,eq,not,true,false} Defined function symbols: {neq} Constructor subsystem: [ eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true ] Rule part & Conj Part: [ eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true, neq(?x,?y) -> not(eq(?x,?y)) ] [ eq(?x,?x) -> true, neq(?x,s(?x)) -> true ] Trying with: R: [ eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true, neq(?x,?y) -> not(eq(?x,?y)) ] E: [ eq(?x,?x) = true, neq(?x,s(?x)) = true ] *** Ground Confluence Check by Rewriting Induction *** Sort: {Nat,Bool} Signature: [ eq : Nat,Nat -> Bool, neq : Nat,Nat -> Bool, not : Bool -> Bool, true : Bool, false : Bool, s : Nat -> Nat, 0 : Nat ] Rule Part: [ eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true, neq(?x,?y) -> not(eq(?x,?y)) ] Conjecture Part: [ eq(?x,?x) = true, neq(?x,s(?x)) = true ] Precedence (by weight): {(0,5),(s,2),(eq,3),(neq,7),(not,6),(true,1),(false,4)} Rule part is confluent. R0 is ground confluent. Check conj part consists of inductive theorems of R0. Rules: [ eq(0,s(?y)) -> false, eq(s(?x),0) -> false, not(true) -> false, not(false) -> true, neq(?x,?y) -> not(eq(?x,?y)) ] Conjectures: [ eq(?x,?x) = true, neq(?x,s(?x)) = true ] STEP 0 ES: [ eq(?x,?x) = true, neq(?x,s(?x)) = true ] HS: [ ] ES0: [ eq(?x,?x) = true, not(eq(?x,s(?x))) = true ] HS0: [ ] ES1: [ eq(?x,?x) = true, not(eq(?x,s(?x))) = true ] HS1: [ ] No equation to expand Failed to prove conj part consists of inductive theorems of R0. new/eq2.trs: Failure(unknown) (15 msec.)