%------------------------------------------------------------------------------ % File : COL060-3 : TPTP v3.3.0. Bugfixed v1.2.0. % Domain : Combinatory Logic % Problem : Find combinator equivalent to Q from B and T % Version : [WM88] (equality) axioms. % Theorem formulation : The combinator is provided and checked. % English : Construct from B and T alone a combinator that behaves as the % combinator Q does, where ((Bx)y)z = x(yz), (Tx)y = yx, % ((Qx)y)z = y(xz). % Refs : [WM88] Wos & McCune (1988), Challenge Problems Focusing on Eq % : [WW+90] Wos et al. (1990), Automated Reasoning Contributes to % Source : [TPTP] % Names : % Status : Unsatisfiable % Rating : 0.12 v3.3.0, 0.00 v2.1.0, 0.29 v2.0.0 % Syntax : Number of clauses : 3 ( 0 non-Horn; 3 unit; 1 RR) % Number of atoms : 3 ( 3 equality) % Maximal clause size : 1 ( 1 average) % Number of predicates : 1 ( 0 propositional; 2-2 arity) % Number of functors : 6 ( 5 constant; 0-2 arity) % Number of variables : 5 ( 0 singleton) % Maximal term depth : 9 ( 4 average) % Comments : % : tptp2X -f tptp:short COL060-3.p % Bugfixes : v1.2.0 : Redundant [fgh]_substitution axioms removed. %------------------------------------------------------------------------------ cnf(b_definition,axiom,( apply(apply(apply(b,X),Y),Z) = apply(X,apply(Y,Z)) )). cnf(t_definition,axiom,( apply(apply(t,X),Y) = apply(Y,X) )). cnf(prove_q_combinator,negated_conjecture,( apply(apply(apply(apply(apply(b,apply(apply(b,apply(t,b)),b)),t),x),y),z) != apply(y,apply(x,z)) )). %------------------------------------------------------------------------------