Logic for Computer Science/Querying Proofs

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Partial Isomorphism[edit]

Definition (partial isomorphism on relational S-Structures)

Let \mathfrak{A} and \mathfrak{B} be relational S-Structures and p be a mapping. Then p is said to be a Partial Isomorphism from \mathfrak{A} to \mathfrak{B} iff all of the following hold

  • dom(p) \subset A and codom(p) \subset B
  • for every a_1, a_2: a_1 = a_2 iff p(a_1) = p(a_2).
  • for every constant symbol c from S and every a: a = c iff p(a) = c
  • for every n-ary relation symbol R from S and a_1, ..., a_n: R^\mathfrak{A} a_1 ... a_n iff R^\mathfrak{B} p(a_1) ... p(a_n)

where a_* \in A, b_* \in B.

Remark

  • I.e. a partial Isomorphism on \mathfrak{A} and \mathfrak{B} is a (plain) Isomprphism on the substructures obtained by choosing dom(p) and codom(p) and 'prune' constants and relations respectively.

Example

  • Let S ={<} be defined over A ={1, 2, 3} and B ={3, 4, 5, 6} the 'usual' way. Then p_1 with 1->3 and 2->4 is a partial isomorphism (since 1 < 2 and p(1) < p(2)) as well as p_3 with 2->3 and 3->6 as well as p_2 with 1->6. Whereas neither p_4 with 1->6 and 2->3 (since 1 < 2 but not p(1) < p(2)) nor p_5 with 2->5 and 3->5 is a partial isomorphism.
  • Let S ={R} with R^\mathfrak{A} =\{(1, 2), (2, 3), (3, 4), (4, 1)\} (a 'square') and R^\mathfrak{B} =\{(1, 2), (1, 3), (3, 4), (3, 5)\} (a 'binary tree'). A ={1, ..., 4} and B ={1, ..., 5}. Then p: {1, 2, 3} -> {1, 3, 4} with 1->1, 2->3, 3->4 is a partial isomorphism from \mathfrak{A} to \mathfrak{B}, since {(1, 2), (2, 3)} becomes {(1, 3), (3, 4)}. Notice that in case R_\mathfrak{B} contained e.g. (4, 1) additionally, p would not be a partial isomorphism.
  • Let S ={<} be defined over A ={1, 2, 3, 4, 5, 6} and B ={1, 3, 5} in the 'usual' way. Then p with 2->3 and 4->5 defines a partial isomorphism. Notice that e.g. \exists_x( x<4 \and 2<x ) holds on \mathfrak{A} but \exists_x( x<p(4) \and p(2)<x ) does not hold on \mathfrak{B}. I.e. in general the validity of sentences with quantifiers is not preserved. So in order to preserve this validity we need sth stronger than partial isomorphism. That is mainly what the Ehrenfeucht-games will provide us with.

Ehrenfeucht-Fraisse Games[edit]

Definition

Suppose that we are given two structures \mathfrak{A} and \mathfrak{B}, each with no function symbols and the same set of relation symbols, and a fixed natural number n.

Then an Ehrenfeucht-Fraisse game is a game with the subsequent properties:

  • Elements
    • a position is a pair of sequences α and β, both of length i, i \in 0, ..., n
    • the starting position is the pair of empty sequences (i = 0)
    • the game is finished iff α and β are both of size n
    • the players are the 'spoiler' S and the 'duplicator' D
  • Moves
    • moves are carried out alternatingly
    • S moves first
    • S chooses exactly one element of either A or B that has not been chosen already (latter is not relevant in the first move)
    • D chooses exactly one element of the other set, i.e. if S has chosen from A then D has to choose from B and if S chose one from B then D has to choose from A.
  • Winning & Information
    • D wins iff the final position defines a partial isomorphism by mapping the i-th elements of α and β, for all i.
    • else S wins
    • both players know both of the structures completely and also know about all the moves played previously


Remark

  • notice that above constants are not mentioned (yet)


Example

  • Let A ={1, 2, 3}, B ={1, 2}, S ={ < }, where '<' is defined the 'usual' way. Here player S can win by playing 2 in the first move and playing the second move corresponding to D's first move in the following way: to 1 he responds 1 and in case of 2 his answer is 3. In both cases D can not follow. He may have isomorphic structures e.g. {2, 3} and {1, 2} but the mapping defined by the sequence of moves is twisted, i.e. it maps 2 to 2 (1st move) and 3 to 1 and thus is not a partial isomorphism.
  • Let A ={1, 2, ..., 9}, B ={1, 2, ..., 8}, S ={ < }, where '<' is defined the 'usual' way. So, what is the best way to play for S? In case S picks 1 from A, D picks 1 from B and the game left is mainly the same as before just with 1 element less but also 1 move played already. But we can make a better deal for the price of one move: by playing 5 in A, D is forced to play 5 (or 4) in B what leaves two possible ways to proceed for S: either playing the {1, 2, 3, 4} vs {1, 2, 3, 4} game what of course would lead to an obvious win for D, or playing the {6, 7, 8, 9} vs {6, 7, 8} game. Choosing the latter the quickest win for S is 7 - 7, 8 - 8 and finally 9 with a 4 move win. Since there is no better way to play for both sides S wins if n > 3 for the structures above.
  • The result from the above example can be generalised to: For linear orders in an n-move Ehrenfeucht game D has a winning strategy if the size of each of A and B is less or equal 2^n.
  • Caption
    The following is an (popular) illustrative representation of an Ehrenfeucht-game: from the two sequences S has won iff the lines connecting matching letters cross (if D can find a match at all), e.g. S wins after three moves:


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Quantifier Rank[edit]

Definition

The function qr(φ) is said to be the Quantifier Ranking of φ iff

  • qr(\varphi) = 0, for φ atomic
  • qr(\varphi_1 \land \varphi_2) = qr(\varphi_1 \lor \varphi_2) = max(qr(\varphi_1), qr(\varphi_2))
  • qr(\lnot \varphi) = qr(\varphi)
  • qr(\forall \varphi) = qr(\exists \varphi) = qr(\varphi)+1

Remark

  • qr(φ) describes the 'depth of quantifier nesting' of φ.
  • We write FO[k] for all Formulas of Quantifier Rank up to k.
  • Notice that in prenex normal form the Quantifier Rank of φ is exactly the number of quantifiers appearing in φ.

Example

  • The sentence \forall_x ( \exists_y ((\lnot x = y) \land x R y ) ) \land ( \exists_y ((\lnot x = z) \land z R x ) ) has the Quantifier Rank 2.
  • The sentence in prenex normal form \forall_x\exists_y\exists_y ((\lnot x = y) \land x R y )  \land ( (\lnot x = z) \land z R x ) has the Quantifier Rank 3. Notice that it is equivalent to the above.

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Ehrenfeucht-Fraisse Theorem[edit]

Theorem

Let \mathfrak{A} and \mathfrak{B} be two structures in a relational vocabulary. Then the following are equivalent

  • \mathfrak{A} and \mathfrak{B} agree on FO[k]
  • \mathfrak{A}\equiv_k \mathfrak{B}

Remark

  • The proof is omitted here
  • The concept of back and forth relations is not mentioned here, since we do not need it subsequently

Expressibility Proofs employing Ehrenfeucht-Fraisse Games[edit]

The basis for considering expressibility by Ehrenfeucht-Fraisse-Games is given by the following corollary from the above theorem:

Corollary

Let P be a property of finite σ-structures. Then the following are equivalent

  • P is not expressible in FO
  • for every k \in \mathbb{N} there exist two finite σ-structures \mathfrak{A}_k and \mathfrak{B}_k, such that the following are both satisfied
    • \mathfrak{A}_k\equiv_k \mathfrak{B}_k
    • \mathfrak{A}_k has P and \mathfrak{B}_k does not have P

Example

  • To begin pick two linear orders say A ={1, 2, 3, 4} and B ={1, 2, 3, 4, 5}. For a two-move Ehrenfeucht game D is to win, obviously. This gives us two structures that satisfy the above conditions for k = 2 and the Property having even cardinality (that A has and B doesn't). Now we have to expand this over all k \in \mathbb{N}. From the above example we adopt that in a linear order of cardinality 2^k or higher D has a winning strategy. Thus we choose the cardinalities depending on k as |A| = 2^k and |B| = 2^k+1. So we have found an even A and an odd B for every k, where D has a winning strategy. Thus (by the corollary) having even/odd cardinality is a property that can not be expressed in FO for finite σ-structures of linear order.
  • Graphs in general are a popular object of interest for FO expressibility. So e.g. it can be shown that the following properties can't be defined in FO:
    • Connectivity
    • Cyclicity
    • Planarity
    • Hamiltonicity
    • n-colorability
    • etc

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