4.93/2.11 YES 4.93/2.11 4.93/2.11 Proof: 4.93/2.11 This system is confluent. 4.93/2.11 Inlined conditions in System R. 4.93/2.11 By \cite{ALS94}, Theorem 4.1. 4.93/2.11 This system is of type 3 or smaller. 4.93/2.11 This system is strongly deterministic. 4.93/2.11 This system is quasi-decreasing. 4.93/2.11 By \cite{O02}, p. 214, Proposition 7.2.50. 4.93/2.11 This system is of type 3 or smaller. 4.93/2.11 This system is deterministic. 4.93/2.11 System R transformed to optimized U(R). 4.93/2.11 This system is terminating. 4.93/2.12 Call external tool: 4.93/2.12 ./ttt2.sh 4.93/2.12 Input: 4.93/2.12 (VAR x y) 4.93/2.12 (RULES 4.93/2.12 add(x, 0) -> x 4.93/2.12 add(x, s(y)) -> s(add(x, y)) 4.93/2.12 quad(x) -> add(add(x, x), add(x, x)) 4.93/2.12 ) 4.93/2.12 4.93/2.12 Matrix Interpretation Processor: dim=1 4.93/2.12 4.93/2.12 interpretation: 4.93/2.12 [quad](x0) = 4x0 + 6, 4.93/2.12 4.93/2.12 [s](x0) = x0, 4.93/2.12 4.93/2.12 [add](x0, x1) = x0 + x1 + 2, 4.93/2.12 4.93/2.12 [0] = 0 4.93/2.12 orientation: 4.93/2.12 add(x,0()) = x + 2 >= x = x 4.93/2.12 4.93/2.12 add(x,s(y)) = x + y + 2 >= x + y + 2 = s(add(x,y)) 4.93/2.12 4.93/2.12 quad(x) = 4x + 6 >= 4x + 6 = add(add(x,x),add(x,x)) 4.93/2.12 problem: 4.93/2.12 add(x,s(y)) -> s(add(x,y)) 4.93/2.12 quad(x) -> add(add(x,x),add(x,x)) 4.93/2.12 Matrix Interpretation Processor: dim=1 4.93/2.12 4.93/2.12 interpretation: 4.93/2.12 [quad](x0) = 4x0 + 1, 4.93/2.12 4.93/2.12 [s](x0) = x0 + 2, 4.93/2.12 4.93/2.12 [add](x0, x1) = x0 + x1 4.93/2.12 orientation: 4.93/2.12 add(x,s(y)) = x + y + 2 >= x + y + 2 = s(add(x,y)) 4.93/2.12 4.93/2.12 quad(x) = 4x + 1 >= 4x = add(add(x,x),add(x,x)) 4.93/2.12 problem: 4.93/2.12 add(x,s(y)) -> s(add(x,y)) 4.93/2.12 Matrix Interpretation Processor: dim=3 4.93/2.12 4.93/2.12 interpretation: 4.93/2.12 [1 0 0] [0] 4.93/2.12 [s](x0) = [0 0 1]x0 + [0] 4.93/2.12 [0 1 1] [1], 4.93/2.12 4.93/2.12 [1 0 0] [1 0 1] 4.93/2.12 [add](x0, x1) = [0 0 0]x0 + [0 1 0]x1 4.93/2.12 [0 0 0] [0 0 1] 4.93/2.12 orientation: 4.93/2.12 [1 0 0] [1 1 1] [1] [1 0 0] [1 0 1] [0] 4.93/2.12 add(x,s(y)) = [0 0 0]x + [0 0 1]y + [0] >= [0 0 0]x + [0 0 1]y + [0] = s(add(x,y)) 4.93/2.12 [0 0 0] [0 1 1] [1] [0 0 0] [0 1 1] [1] 4.93/2.12 problem: 4.93/2.12 4.93/2.12 Qed 4.93/2.12 All 0 critical pairs are joinable. 4.93/2.12 4.93/2.15 EOF