TSTP Solution File: SYN083+1 by Bliksem---1.12

View Problem - Process Solution

%------------------------------------------------------------------------------
% File     : Bliksem---1.12
% Problem  : SYN083+1 : TPTP v8.1.0. Released v2.0.0.
% Transfm  : none
% Format   : tptp:raw
% Command  : bliksem %s

% Computer : n017.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  : 0s
% DateTime : Thu Jul 21 02:47:35 EDT 2022

% Result   : Theorem 0.42s 1.06s
% Output   : Refutation 0.42s
% Verified : 
% SZS Type : -

% Comments : 
%------------------------------------------------------------------------------
%----WARNING: Could not form TPTP format derivation
%------------------------------------------------------------------------------
%----ORIGINAL SYSTEM OUTPUT
% 0.00/0.11  % Problem  : SYN083+1 : TPTP v8.1.0. Released v2.0.0.
% 0.00/0.12  % Command  : bliksem %s
% 0.12/0.33  % Computer : n017.cluster.edu
% 0.12/0.33  % Model    : x86_64 x86_64
% 0.12/0.33  % CPU      : Intel(R) Xeon(R) CPU E5-2620 v4 @ 2.10GHz
% 0.12/0.33  % Memory   : 8042.1875MB
% 0.12/0.33  % OS       : Linux 3.10.0-693.el7.x86_64
% 0.12/0.33  % CPULimit : 300
% 0.12/0.33  % DateTime : Tue Jul 12 08:12:04 EDT 2022
% 0.12/0.33  % CPUTime  : 
% 0.42/1.06  *** allocated 10000 integers for termspace/termends
% 0.42/1.06  *** allocated 10000 integers for clauses
% 0.42/1.06  *** allocated 10000 integers for justifications
% 0.42/1.06  Bliksem 1.12
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Automatic Strategy Selection
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Clauses:
% 0.42/1.06  
% 0.42/1.06  { f( X, f( Y, Z ) ) = f( f( X, Y ), Z ) }.
% 0.42/1.06  { ! f( skol1, f( skol2, f( skol3, skol4 ) ) ) = f( f( f( skol1, skol2 ), 
% 0.42/1.06    skol3 ), skol4 ) }.
% 0.42/1.06  
% 0.42/1.06  percentage equality = 1.000000, percentage horn = 1.000000
% 0.42/1.06  This is a pure equality problem
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Options Used:
% 0.42/1.06  
% 0.42/1.06  useres =            1
% 0.42/1.06  useparamod =        1
% 0.42/1.06  useeqrefl =         1
% 0.42/1.06  useeqfact =         1
% 0.42/1.06  usefactor =         1
% 0.42/1.06  usesimpsplitting =  0
% 0.42/1.06  usesimpdemod =      5
% 0.42/1.06  usesimpres =        3
% 0.42/1.06  
% 0.42/1.06  resimpinuse      =  1000
% 0.42/1.06  resimpclauses =     20000
% 0.42/1.06  substype =          eqrewr
% 0.42/1.06  backwardsubs =      1
% 0.42/1.06  selectoldest =      5
% 0.42/1.06  
% 0.42/1.06  litorderings [0] =  split
% 0.42/1.06  litorderings [1] =  extend the termordering, first sorting on arguments
% 0.42/1.06  
% 0.42/1.06  termordering =      kbo
% 0.42/1.06  
% 0.42/1.06  litapriori =        0
% 0.42/1.06  termapriori =       1
% 0.42/1.06  litaposteriori =    0
% 0.42/1.06  termaposteriori =   0
% 0.42/1.06  demodaposteriori =  0
% 0.42/1.06  ordereqreflfact =   0
% 0.42/1.06  
% 0.42/1.06  litselect =         negord
% 0.42/1.06  
% 0.42/1.06  maxweight =         15
% 0.42/1.06  maxdepth =          30000
% 0.42/1.06  maxlength =         115
% 0.42/1.06  maxnrvars =         195
% 0.42/1.06  excuselevel =       1
% 0.42/1.06  increasemaxweight = 1
% 0.42/1.06  
% 0.42/1.06  maxselected =       10000000
% 0.42/1.06  maxnrclauses =      10000000
% 0.42/1.06  
% 0.42/1.06  showgenerated =    0
% 0.42/1.06  showkept =         0
% 0.42/1.06  showselected =     0
% 0.42/1.06  showdeleted =      0
% 0.42/1.06  showresimp =       1
% 0.42/1.06  showstatus =       2000
% 0.42/1.06  
% 0.42/1.06  prologoutput =     0
% 0.42/1.06  nrgoals =          5000000
% 0.42/1.06  totalproof =       1
% 0.42/1.06  
% 0.42/1.06  Symbols occurring in the translation:
% 0.42/1.06  
% 0.42/1.06  {}  [0, 0]      (w:1, o:2, a:1, s:1, b:0), 
% 0.42/1.06  .  [1, 2]      (w:1, o:19, a:1, s:1, b:0), 
% 0.42/1.06  !  [4, 1]      (w:0, o:14, a:1, s:1, b:0), 
% 0.42/1.06  =  [13, 2]      (w:1, o:0, a:0, s:1, b:0), 
% 0.42/1.06  ==>  [14, 2]      (w:1, o:0, a:0, s:1, b:0), 
% 0.42/1.06  f  [38, 2]      (w:1, o:43, a:1, s:1, b:0), 
% 0.42/1.06  skol1  [40, 0]      (w:1, o:10, a:1, s:1, b:1), 
% 0.42/1.06  skol2  [41, 0]      (w:1, o:11, a:1, s:1, b:1), 
% 0.42/1.06  skol3  [42, 0]      (w:1, o:12, a:1, s:1, b:1), 
% 0.42/1.06  skol4  [43, 0]      (w:1, o:13, a:1, s:1, b:1).
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Starting Search:
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Bliksems!, er is een bewijs:
% 0.42/1.06  % SZS status Theorem
% 0.42/1.06  % SZS output start Refutation
% 0.42/1.06  
% 0.42/1.06  (0) {G0,W11,D4,L1,V3,M1} I { f( X, f( Y, Z ) ) ==> f( f( X, Y ), Z ) }.
% 0.42/1.06  (1) {G1,W0,D0,L0,V0,M0} I;d(0);d(0);q {  }.
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  % SZS output end Refutation
% 0.42/1.06  found a proof!
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Unprocessed initial clauses:
% 0.42/1.06  
% 0.42/1.06  (3) {G0,W11,D4,L1,V3,M1}  { f( X, f( Y, Z ) ) = f( f( X, Y ), Z ) }.
% 0.42/1.06  (4) {G0,W15,D5,L1,V0,M1}  { ! f( skol1, f( skol2, f( skol3, skol4 ) ) ) = f
% 0.42/1.06    ( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Total Proof:
% 0.42/1.06  
% 0.42/1.06  subsumption: (0) {G0,W11,D4,L1,V3,M1} I { f( X, f( Y, Z ) ) ==> f( f( X, Y
% 0.42/1.06     ), Z ) }.
% 0.42/1.06  parent0: (3) {G0,W11,D4,L1,V3,M1}  { f( X, f( Y, Z ) ) = f( f( X, Y ), Z )
% 0.42/1.06     }.
% 0.42/1.06  substitution0:
% 0.42/1.06     X := X
% 0.42/1.06     Y := Y
% 0.42/1.06     Z := Z
% 0.42/1.06  end
% 0.42/1.06  permutation0:
% 0.42/1.06     0 ==> 0
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  paramod: (15) {G1,W15,D5,L1,V0,M1}  { ! f( skol1, f( f( skol2, skol3 ), 
% 0.42/1.06    skol4 ) ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  parent0[0]: (0) {G0,W11,D4,L1,V3,M1} I { f( X, f( Y, Z ) ) ==> f( f( X, Y )
% 0.42/1.06    , Z ) }.
% 0.42/1.06  parent1[0; 4]: (4) {G0,W15,D5,L1,V0,M1}  { ! f( skol1, f( skol2, f( skol3, 
% 0.42/1.06    skol4 ) ) ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  substitution0:
% 0.42/1.06     X := skol2
% 0.42/1.06     Y := skol3
% 0.42/1.06     Z := skol4
% 0.42/1.06  end
% 0.42/1.06  substitution1:
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  paramod: (18) {G1,W15,D5,L1,V0,M1}  { ! f( f( skol1, f( skol2, skol3 ) ), 
% 0.42/1.06    skol4 ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  parent0[0]: (0) {G0,W11,D4,L1,V3,M1} I { f( X, f( Y, Z ) ) ==> f( f( X, Y )
% 0.42/1.06    , Z ) }.
% 0.42/1.06  parent1[0; 2]: (15) {G1,W15,D5,L1,V0,M1}  { ! f( skol1, f( f( skol2, skol3
% 0.42/1.06     ), skol4 ) ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  substitution0:
% 0.42/1.06     X := skol1
% 0.42/1.06     Y := f( skol2, skol3 )
% 0.42/1.06     Z := skol4
% 0.42/1.06  end
% 0.42/1.06  substitution1:
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  paramod: (19) {G1,W15,D5,L1,V0,M1}  { ! f( f( f( skol1, skol2 ), skol3 ), 
% 0.42/1.06    skol4 ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  parent0[0]: (0) {G0,W11,D4,L1,V3,M1} I { f( X, f( Y, Z ) ) ==> f( f( X, Y )
% 0.42/1.06    , Z ) }.
% 0.42/1.06  parent1[0; 3]: (18) {G1,W15,D5,L1,V0,M1}  { ! f( f( skol1, f( skol2, skol3
% 0.42/1.06     ) ), skol4 ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  substitution0:
% 0.42/1.06     X := skol1
% 0.42/1.06     Y := skol2
% 0.42/1.06     Z := skol3
% 0.42/1.06  end
% 0.42/1.06  substitution1:
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  eqrefl: (20) {G0,W0,D0,L0,V0,M0}  {  }.
% 0.42/1.06  parent0[0]: (19) {G1,W15,D5,L1,V0,M1}  { ! f( f( f( skol1, skol2 ), skol3 )
% 0.42/1.06    , skol4 ) = f( f( f( skol1, skol2 ), skol3 ), skol4 ) }.
% 0.42/1.06  substitution0:
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  subsumption: (1) {G1,W0,D0,L0,V0,M0} I;d(0);d(0);q {  }.
% 0.42/1.06  parent0: (20) {G0,W0,D0,L0,V0,M0}  {  }.
% 0.42/1.06  substitution0:
% 0.42/1.06  end
% 0.42/1.06  permutation0:
% 0.42/1.06  end
% 0.42/1.06  
% 0.42/1.06  Proof check complete!
% 0.42/1.06  
% 0.42/1.06  Memory use:
% 0.42/1.06  
% 0.42/1.06  space for terms:        84
% 0.42/1.06  space for clauses:      178
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  clauses generated:      2
% 0.42/1.06  clauses kept:           2
% 0.42/1.06  clauses selected:       0
% 0.42/1.06  clauses deleted:        0
% 0.42/1.06  clauses inuse deleted:  0
% 0.42/1.06  
% 0.42/1.06  subsentry:          33
% 0.42/1.06  literals s-matched: 10
% 0.42/1.06  literals matched:   10
% 0.42/1.06  full subsumption:   0
% 0.42/1.06  
% 0.42/1.06  checksum:           -50342155
% 0.42/1.06  
% 0.42/1.06  
% 0.42/1.06  Bliksem ended
%------------------------------------------------------------------------------