The rewrite relation of the following TRS is considered.
not(x) | → | xor(x,true) | (1) |
implies(x,y) | → | xor(and(x,y),xor(x,true)) | (2) |
or(x,y) | → | xor(and(x,y),xor(x,y)) | (3) |
=(x,y) | → | xor(x,xor(y,true)) | (4) |
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not#(z0) |
implies#(z0,z1) |
or#(z0,z1) |
=#(z0,z1) |
not(z0) | → | xor(z0,true) | (5) |
implies(z0,z1) | → | xor(and(z0,z1),xor(z0,true)) | (7) |
or(z0,z1) | → | xor(and(z0,z1),xor(z0,z1)) | (9) |
=(z0,z1) | → | xor(z0,xor(z1,true)) | (11) |
implies#(z0,z1) | → | c1 | (8) |
=#(z0,z1) | → | c3 | (12) |
[c] | = | 0 |
[c1] | = | 0 |
[c2] | = | 0 |
[c3] | = | 0 |
[not#(x1)] | = | 0 |
[implies#(x1, x2)] | = | 1 |
[or#(x1, x2)] | = | 0 |
[=#(x1, x2)] | = | 1 |
not#(z0) | → | c | (6) |
implies#(z0,z1) | → | c1 | (8) |
or#(z0,z1) | → | c2 | (10) |
=#(z0,z1) | → | c3 | (12) |
or#(z0,z1) | → | c2 | (10) |
[c] | = | 0 |
[c1] | = | 0 |
[c2] | = | 0 |
[c3] | = | 0 |
[not#(x1)] | = | 0 |
[implies#(x1, x2)] | = | 0 |
[or#(x1, x2)] | = | 1 |
[=#(x1, x2)] | = | 0 |
not#(z0) | → | c | (6) |
implies#(z0,z1) | → | c1 | (8) |
or#(z0,z1) | → | c2 | (10) |
=#(z0,z1) | → | c3 | (12) |
not#(z0) | → | c | (6) |
[c] | = | 0 |
[c1] | = | 0 |
[c2] | = | 0 |
[c3] | = | 0 |
[not#(x1)] | = | 1 |
[implies#(x1, x2)] | = | 0 |
[or#(x1, x2)] | = | 0 |
[=#(x1, x2)] | = | 0 |
not#(z0) | → | c | (6) |
implies#(z0,z1) | → | c1 | (8) |
or#(z0,z1) | → | c2 | (10) |
=#(z0,z1) | → | c3 | (12) |
There are no rules in the TRS R. Hence, R/S has complexity O(1).