Logic for Computer Science/Querying Proofs

Partial Isomorphism
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. $$\exists_x( x<4 \land 2<x )$$ holds on $$\mathfrak{A}$$ but $$\exists_x( x<p(4) \land p(2)<x )$$ 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
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.


 * [[Image:Logic-FMT-EFGame-Strg01.gif|thumb|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
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
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
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