reserve k,n,m for Nat,
  a,x,X,Y for set,
  D,D1,D2,S for non empty set,
  p,q for FinSequence of NAT;
reserve F,F1,G,G1,H,H1,H2 for LTL-formula;
reserve sq,sq9 for FinSequence;
reserve L,L9 for FinSequence;
reserve j for Nat;
reserve j1 for Element of NAT;

theorem Th37:
  (G is negative or G is next) & F is_proper_subformula_of G
  implies F is_subformula_of (the_argument_of G)
proof
  assume that
A1: G is negative or G is next and
A2: F is_subformula_of G and
A3: F <> G;
  consider n,L such that
A4: 1 <= n and
A5: len L = n and
A6: L.1 = F and
A7: L.n = G and
A8: for k st 1 <= k & k < n ex H1,F1 st L.k = H1 & L.(k + 1) = F1 & H1
  is_immediate_constituent_of F1 by A2;
  1 < n by A3,A4,A6,A7,XXREAL_0:1;
  then 1 + 1 <= n by NAT_1:13;
  then consider k be Nat such that
A9: n = 2 + k by NAT_1:10;
  reconsider L1 = L|(Seg(1 + k)) as FinSequence by FINSEQ_1:15;
  take m = 1 + k, L1;
  thus
A10: 1 <= m by NAT_1:11;
  1 + k <= 1 + k + 1 by NAT_1:11;
  hence len L1 = m by A5,A9,FINSEQ_1:17;
A11: now
    let j;
    assume 1 <= j & j <= m;
    then j in { j1 where j1 is Nat : 1 <= j1 & j1 <= 1 + k };
    then j in Seg(1 + k) by FINSEQ_1:def 1;
    hence L1.j = L.j by FUNCT_1:49;
  end;
  hence L1.1 = F by A6,A10;
  m < m + 1 by NAT_1:13;
  then consider F1,G1 such that
A12: L.m = F1 and
A13: L.(m + 1) = G1 & F1 is_immediate_constituent_of G1 by A8,A9,NAT_1:11;
  F1 = (the_argument_of G) by A1,A7,A9,A13,Th20,Th21;
  hence L1.m = (the_argument_of G) by A10,A11,A12;
  let j;
  assume that
A14: 1 <= j and
A15: j < m;
  m <= m + 1 by NAT_1:11;
  then j < n by A9,A15,XXREAL_0:2;
  then consider F1,G1 such that
A16: L.j = F1 & L.(j + 1) = G1 & F1 is_immediate_constituent_of G1 by A8,A14;
  take F1,G1;
  1 <= 1 + j & j + 1 <= m by A14,A15,NAT_1:13;
  hence thesis by A11,A14,A15,A16;
end;
