TSTP Solution File: SET108+1 by Twee---2.4.2
View Problem
- Process Solution
%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SET108+1 : TPTP v8.1.2. Bugfixed v5.4.0.
% Transfm : none
% Format : tptp:raw
% Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% Computer : n016.cluster.edu
% Model : x86_64 x86_64
% CPU : Intel(R) Xeon(R) CPU E5-2620 v4 2.10GHz
% Memory : 8042.1875MB
% OS : Linux 3.10.0-693.el7.x86_64
% CPULimit : 300s
% WCLimit : 300s
% DateTime : Thu Aug 31 15:31:18 EDT 2023
% Result : Theorem 9.93s 1.63s
% Output : Proof 9.93s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.12 % Problem : SET108+1 : TPTP v8.1.2. Bugfixed v5.4.0.
% 0.00/0.13 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.13/0.34 % Computer : n016.cluster.edu
% 0.13/0.34 % Model : x86_64 x86_64
% 0.13/0.34 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.13/0.34 % Memory : 8042.1875MB
% 0.13/0.34 % OS : Linux 3.10.0-693.el7.x86_64
% 0.13/0.34 % CPULimit : 300
% 0.13/0.34 % WCLimit : 300
% 0.13/0.34 % DateTime : Sat Aug 26 16:54:27 EDT 2023
% 0.13/0.34 % CPUTime :
% 9.93/1.63 Command-line arguments: --no-flatten-goal
% 9.93/1.63
% 9.93/1.63 % SZS status Theorem
% 9.93/1.63
% 9.93/1.63 % SZS output start Proof
% 9.93/1.63 Take the following subset of the input axioms:
% 9.93/1.63 fof(complement, axiom, ![X, Z]: (member(Z, complement(X)) <=> (member(Z, universal_class) & ~member(Z, X)))).
% 9.93/1.63 fof(disjoint_defn, axiom, ![Y, X2]: (disjoint(X2, Y) <=> ![U]: ~(member(U, X2) & member(U, Y)))).
% 9.93/1.63 fof(domain_of, axiom, ![X2, Z2]: (member(Z2, domain_of(X2)) <=> (member(Z2, universal_class) & restrict(X2, singleton(Z2), universal_class)!=null_class))).
% 9.93/1.63 fof(existence_of_first_and_second, conjecture, ![X2]: ?[V, U2]: ((member(U2, universal_class) & (member(V, universal_class) & X2=ordered_pair(U2, V))) | (~?[Y2, Z2]: (member(Y2, universal_class) & (member(Z2, universal_class) & X2=ordered_pair(Y2, Z2))) & (U2=X2 & V=X2)))).
% 9.93/1.63 fof(null_class_defn, axiom, ![X2]: ~member(X2, null_class)).
% 9.93/1.63
% 9.93/1.63 Now clausify the problem and encode Horn clauses using encoding 3 of
% 9.93/1.63 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 9.93/1.63 We repeatedly replace C & s=t => u=v by the two clauses:
% 9.93/1.63 fresh(y, y, x1...xn) = u
% 9.93/1.63 C => fresh(s, t, x1...xn) = v
% 9.93/1.63 where fresh is a fresh function symbol and x1..xn are the free
% 9.93/1.63 variables of u and v.
% 9.93/1.64 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 9.93/1.64 input problem has no model of domain size 1).
% 9.93/1.64
% 9.93/1.64 The encoding turns the above axioms into the following unit equations and goals:
% 9.93/1.64
% 9.93/1.64 Axiom 1 (existence_of_first_and_second): fresh61(X, X) = x.
% 9.93/1.64 Axiom 2 (existence_of_first_and_second_1): fresh59(X, X) = true2.
% 9.93/1.64 Axiom 3 (existence_of_first_and_second_2): fresh57(X, X) = true2.
% 9.93/1.64 Axiom 4 (existence_of_first_and_second): fresh62(X, X, Y) = ordered_pair(y, z).
% 9.93/1.64 Axiom 5 (existence_of_first_and_second): fresh62(X, x, Y) = fresh61(Y, x).
% 9.93/1.64 Axiom 6 (existence_of_first_and_second_1): fresh60(X, X, Y) = member(z, universal_class).
% 9.93/1.64 Axiom 7 (existence_of_first_and_second_1): fresh60(X, x, Y) = fresh59(Y, x).
% 9.93/1.64 Axiom 8 (existence_of_first_and_second_2): fresh58(X, X, Y) = member(y, universal_class).
% 9.93/1.64 Axiom 9 (existence_of_first_and_second_2): fresh58(X, x, Y) = fresh57(Y, x).
% 9.93/1.64
% 9.93/1.64 Goal 1 (existence_of_first_and_second_3): tuple3(x, member(X, universal_class), member(Y, universal_class)) = tuple3(ordered_pair(X, Y), true2, true2).
% 9.93/1.64 The goal is true when:
% 9.93/1.64 X = y
% 9.93/1.64 Y = z
% 9.93/1.64
% 9.93/1.64 Proof:
% 9.93/1.64 tuple3(x, member(y, universal_class), member(z, universal_class))
% 9.93/1.64 = { by axiom 8 (existence_of_first_and_second_2) R->L }
% 9.93/1.64 tuple3(x, fresh58(x, x, x), member(z, universal_class))
% 9.93/1.64 = { by axiom 9 (existence_of_first_and_second_2) }
% 9.93/1.64 tuple3(x, fresh57(x, x), member(z, universal_class))
% 9.93/1.64 = { by axiom 3 (existence_of_first_and_second_2) }
% 9.93/1.64 tuple3(x, true2, member(z, universal_class))
% 9.93/1.64 = { by axiom 6 (existence_of_first_and_second_1) R->L }
% 9.93/1.64 tuple3(x, true2, fresh60(x, x, x))
% 9.93/1.64 = { by axiom 7 (existence_of_first_and_second_1) }
% 9.93/1.64 tuple3(x, true2, fresh59(x, x))
% 9.93/1.64 = { by axiom 2 (existence_of_first_and_second_1) }
% 9.93/1.64 tuple3(x, true2, true2)
% 9.93/1.64 = { by axiom 1 (existence_of_first_and_second) R->L }
% 9.93/1.64 tuple3(fresh61(x, x), true2, true2)
% 9.93/1.64 = { by axiom 5 (existence_of_first_and_second) R->L }
% 9.93/1.64 tuple3(fresh62(x, x, x), true2, true2)
% 9.93/1.64 = { by axiom 4 (existence_of_first_and_second) }
% 9.93/1.64 tuple3(ordered_pair(y, z), true2, true2)
% 9.93/1.64 % SZS output end Proof
% 9.93/1.64
% 9.93/1.64 RESULT: Theorem (the conjecture is true).
%------------------------------------------------------------------------------