TSTP Solution File: SWW478+2 by Twee---2.4.2
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%------------------------------------------------------------------------------
% File : Twee---2.4.2
% Problem : SWW478+2 : TPTP v8.1.2. Released v5.3.0.
% 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 : n015.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 : Fri Sep 1 00:55:20 EDT 2023
% Result : Theorem 37.98s 5.31s
% Output : Proof 37.98s
% Verified :
% SZS Type : -
% Comments :
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.08/0.14 % Problem : SWW478+2 : TPTP v8.1.2. Released v5.3.0.
% 0.08/0.15 % Command : parallel-twee %s --tstp --conditional-encoding if --smaller --drop-non-horn --give-up-on-saturation --explain-encoding --formal-proof
% 0.14/0.36 % Computer : n015.cluster.edu
% 0.14/0.36 % Model : x86_64 x86_64
% 0.14/0.36 % CPU : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.14/0.36 % Memory : 8042.1875MB
% 0.14/0.36 % OS : Linux 3.10.0-693.el7.x86_64
% 0.14/0.36 % CPULimit : 300
% 0.14/0.36 % WCLimit : 300
% 0.14/0.36 % DateTime : Sun Aug 27 22:22:39 EDT 2023
% 0.14/0.36 % CPUTime :
% 37.98/5.31 Command-line arguments: --kbo-weight0 --lhs-weight 5 --flip-ordering --normalise-queue-percent 10 --cp-renormalise-threshold 10 --goal-heuristic
% 37.98/5.31
% 37.98/5.31 % SZS status Theorem
% 37.98/5.31
% 37.98/5.32 % SZS output start Proof
% 37.98/5.32 Take the following subset of the input axioms:
% 37.98/5.32 fof(conj_0, conjecture, hBOOL(member773094996on_val(hAPP_P1886180715on_val(hAPP_P1870962205on_val(produc1441475159on_val, hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, ea), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, ha), fun_up1149430426on_val(la, v_1, hAPP_val_option_val(some_val, v))))), hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, e_a), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, h_a), l_a))), red(p)))).
% 37.98/5.32 fof(fact_1_InitBlockRed_I1_J, axiom, hBOOL(member773094996on_val(hAPP_P1886180715on_val(hAPP_P1870962205on_val(produc1441475159on_val, hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, ea), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, ha), fun_up1149430426on_val(la, v_1, hAPP_val_option_val(some_val, v))))), hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, e_a), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, h_a), l_a))), red(p)))).
% 37.98/5.32
% 37.98/5.32 Now clausify the problem and encode Horn clauses using encoding 3 of
% 37.98/5.32 http://www.cse.chalmers.se/~nicsma/papers/horn.pdf.
% 37.98/5.32 We repeatedly replace C & s=t => u=v by the two clauses:
% 37.98/5.32 fresh(y, y, x1...xn) = u
% 37.98/5.32 C => fresh(s, t, x1...xn) = v
% 37.98/5.32 where fresh is a fresh function symbol and x1..xn are the free
% 37.98/5.32 variables of u and v.
% 37.98/5.32 A predicate p(X) is encoded as p(X)=true (this is sound, because the
% 37.98/5.32 input problem has no model of domain size 1).
% 37.98/5.32
% 37.98/5.32 The encoding turns the above axioms into the following unit equations and goals:
% 37.98/5.32
% 37.98/5.32 Axiom 1 (fact_1_InitBlockRed_I1_J): hBOOL(member773094996on_val(hAPP_P1886180715on_val(hAPP_P1870962205on_val(produc1441475159on_val, hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, ea), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, ha), fun_up1149430426on_val(la, v_1, hAPP_val_option_val(some_val, v))))), hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, e_a), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, h_a), l_a))), red(p))) = true2.
% 37.98/5.32
% 37.98/5.32 Goal 1 (conj_0): hBOOL(member773094996on_val(hAPP_P1886180715on_val(hAPP_P1870962205on_val(produc1441475159on_val, hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, ea), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, ha), fun_up1149430426on_val(la, v_1, hAPP_val_option_val(some_val, v))))), hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, e_a), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, h_a), l_a))), red(p))) = true2.
% 37.98/5.32 Proof:
% 37.98/5.32 hBOOL(member773094996on_val(hAPP_P1886180715on_val(hAPP_P1870962205on_val(produc1441475159on_val, hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, ea), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, ha), fun_up1149430426on_val(la, v_1, hAPP_val_option_val(some_val, v))))), hAPP_P604205461on_val(hAPP_e1659493427on_val(produc1259058957on_val, e_a), hAPP_f1727192346on_val(hAPP_f1849790461on_val(produc899768717on_val, h_a), l_a))), red(p)))
% 37.98/5.32 = { by axiom 1 (fact_1_InitBlockRed_I1_J) }
% 37.98/5.32 true2
% 37.98/5.32 % SZS output end Proof
% 37.98/5.32
% 37.98/5.32 RESULT: Theorem (the conjecture is true).
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