Unit4_CL_Unit_4_on Computation Logic_srm

sanshansh 11 views 55 slides Jun 07, 2024
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About This Presentation

Computational Logic


Slide Content

Computational Logic UNIT4(First Order logic) Mr. Naresh Sharma Assistant Professor Department of CSE SRMIST/NCR

Syllabus (proofs in predicate logic) Axiomatic System FC Introduction Examples Theorems Monotonicity Deduction, RA, Fitness, Paradox of material Implication, Strong Generalization Adequacy of FC to FL Compactness of FL Laws of FL Natural Deduction Analytic Tableaux

Axiomatic system

terminologies Finite sequence of formulas Proof Axiom Derived by application of some inference rule on earlier formula Theorem ⊢ X X is provable Σ ⊢ Y Last formula of proof  ⊢ FC X. A set of formulas Σ is said to be inconsistent if there exists a formula Y such that Σ ⊢ Y and Σ ⊢ ¬ Y , else Σ is said to be consistent . Proof of the consequence Consequence is provable Proof System First order Calculus (FC) Sequence of formulas

Axiom schemes of FC (A1) X → (Y → X) (A2) (X → (Y → Z)) → ((X → Y ) → (X → Z)) (A3) (¬X → ¬Y ) → ((¬X → Y ) → X) (A4) ∀ xY → Y [x/t], provided x is free for t in Y. (A5) ∀x(Y → Z) → (Y → ∀ xZ ), provided x does not occur free in Y. (A6) (t ≈ t) (A7) ((s ≈ t) → (X[x/s] → X[x/t]), provided x is free for s, t in X. Let X, Y, Z, be Formulas x be variable and s, t be terms Axiom schemes of PC Equality predicate Semantic counterpart of ⊨∀X → X[x/t]. Quantifiers

Definitions and rules (D1) p ∧ q ≐ ¬(p → ¬q) (D2) p ∨ q ≐ ¬p → q (D3) p ↔ q ≐ ¬((p → q) → ¬(q → p)) (D4) ⊤ ≐ p → p (D5) ⊥ ≐ ¬(p → p) (D6) ∃ xX ≐ ¬∀ x¬X Single formula Two formulas Inference Provided x is not free in any premise used thus far.

exercises ∀ x∀yX ⊢ ∀ y∀xX ⊢ ∀ xX → ∀ xX – different from the one we discussed If x does not occur free in X, then ⊢ X → ∀ xX . ∀ x¬X ⊢ ¬∀ xX ⊢ (s ≈ t) ⊢ (t ≈ s).

∀ x∀yX ⊢ ∀ y∀xX ⊢ ∀ xX → ∀ xX

If x does not occur free in X, then ⊢ X → ∀ xX X → X P ∀x(X → X) UG ∀x(X → X) → (X → ∀ xX ) A5 X → ∀ xX MP ∀ x¬X ⊢ ¬∀ xX ∀ x¬X P ∀ x¬X → ¬X A4  1. ¬X MP ∀ xX → X A4 (∀ xX → X) → (¬X → ¬∀ xX ) Th ¬X → ¬∀ xX MP ¬∀ xX 1, MP ⊢ (s ≈ t) ⊢ (t ≈ s) (s ≈ t) P (s ≈ t) → ((s ≈ s) → (t ≈ s) A7,X=(x ≈ s) (s ≈ s) → (t ≈ s) MP (s ≈ s) A6 (t ≈ s) MP

theorems First Order Logic

RAA Same as that in PC Let Σ be a set of formulas and let X be a formula. Σ ⊨ X iff Σ ∪ {¬ X} is inconsistent. Σ ⊨ ¬ X iff Σ ∪ { X} is inconsistent If i is a model of Σ , then as Σ ⊨ w, i ( ¬ w) =0. If i ⊭ Σ , then i ⊭ x for some x∈ Σ ; hence i ⊭ Σ ∪{¬w} If i is not a model of Σ , then i ⊭ Σ ∪{¬w} Thus Σ ∪{¬w} is unsatisfiable. Conversely Let Σ ∪{¬w} be unsatisfiable and i ⊨ Σ , then i (¬w)=0, hence i ⊨ w, Therefore Σ ⊨ w

monotonocity Same as that in PC Let Σ , Γ be set of formulas and X a formula. Suppose that Σ ⊆ Γ If Σ ⊨ X, then Γ ⊨ X. If Σ ⊨ p and i ⊨ Γ , then i (x)=1 for every x ∈ Γ . As Σ ⊆ Γ , i (y)=1 for every y ∈ Σ , i ⊨ Σ . Since Σ ⊨ p, i (p)=1. Therefore Γ ⊨ p

Deduction Let us assume that there is a proof P whose last formula is X → Y then let us prove Σ ∪ { X} ⊢ Y. Let Σ be a set of formulas, and let X, Y be formulas. Then, Σ ⊨ X → Y iff Σ ∪ { X} ⊨ Y X → Y Σ ⊢ X → Y X P in Σ ∪ { X} Y MP Let us assume that there is a proof P whose last formula is Σ ∪ { X} ⊢ Y then let us prove X → Y If P has I formula; then it is Y ➾ axiom, Premise in Σ or X itself Y → (X → Y ) A1 Y Axiom / Premise X → Y MP ⊢ X → X PC Theorem If Σ ⊨ X, then Γ ⊨ X Monotonocity Σ ⊢ X → X

Proof contd … Induction hypothesis: If P has number of formulas to be less than n, then there is a proof of Σ ⊢ X → Y. Suppose that Σ ∪ { X} ⊢ Y has a proof P1 of n formulas. Then the n th formula is Y. Y can be axiom, Premise in Σ or X itself Derived from two earlier formulas using MP Derived from an earlier formula by UG Covered in base case Y → (X → Y ) A1 Y Axiom / Premise X → Y MP ⊢ X → X PC Theorem If Σ ⊨ X, then Γ ⊨ X Monotonocity Σ ⊢ X → X

Proof contd … Derived from two earlier formulas using MP m steps to derive Z, k steps to derive Z  Y. At step n using steps m and m+k apply MP to derive Y P1: 1. m. Z m+k . (Z → Y ) MP n. Y m, m+k , MP

Proof contd … P5: 1. … . . . i . X  Z . . . i+j . X → (Z → Y ) i+j+1 (X → (Z → Y )) → ((X → Z) → (X → Y )) A2 i+j+2 (X → Z) → (X → Y ) MP i+j+3 X → Y MP P2 ( i steps): Σ ⊢ X → Z, P3 (j steps): Σ ⊢ X → (Z → Y )

Proof contd … m steps to derive A. At step n using steps m and n-1 apply UG to derive ∀x A As Y = ∀ xA , P5 is a proof for Σ ⊢ X → Y.

example ⊢ ∀x(X → Y ) → (∀ x¬Y → ∀ x¬X ) Deduction Theorem {∀x(X → Y ), (∀ x¬Y ) } ⊢ ∀ x¬X ∀x(X → Y ) P ∀x (X → Y ) →(X → Y ) A4 X → Y MP (X → Y ) → (¬Y →¬X) Th (¬Y →¬X) MP ∀ x¬Y P ∀ x¬Y → ¬Y A4 ¬Y MP ¬X 1,MP ∀ x¬X UG Back

example ⊢ ∀x(X → Y ) → (¬∀ x¬X → ¬∀ x¬Y ). {∀x(X → Y ), (∀ x¬Y ) } ⊢ ∀ x¬X Deduction Theorem ∀x(X → Y ) ⊢ (∀ x¬Y ) → (∀ x¬X ) Contraposition ⊢ (∀ x¬Y → ∀ x¬X ) → (¬∀ x¬X → ¬∀ x¬Y ) MP ∀x(X → Y ) ⊢ ¬∀ x¬X → ¬∀ x¬Y Deduction Theorem ⊢ ∀x(X → Y ) → (¬∀ x¬X → ¬∀ x¬Y )

Exercises (RAA & Deduction) ⊢ ∀x(X → Y ) → (¬∀ x¬X → ¬∀ x¬Y ) If x is not free in Y , then ⊢ ¬(∀ xX → Y ) → ∀x¬(X → Y ) ⊢ ∀x((x ≈ f(y)) → Qx ) → Qf (y). {Pa, ∀x( Px → Qx ), ∀x(Rx → ¬ Qx ), Rb } ⊢ ¬(a ≈ b). ∀ x∀y (f(x, y) ≈ f(y, x)), ∀ x∀y (f(x, y) ≈ y) ⊨ ¬∀x¬∀y(x ≈ y)

If x is not free in Y , then ⊢ ¬(∀ xX → Y ) → ∀x¬(X → Y ) ⊢ ¬(∀ xX → Y ) → ∀x¬(X → Y ) iff ¬(∀ xX → Y ) ⊢ ∀x¬(X → Y ) iff {¬(∀ xX → Y ), ¬∀x¬(X → Y )} is inconsistent. iff ¬∀x¬(X → Y ) ⊢ ∀ xX → Y iff ¬∀x¬(X → Y ), ∀ xX ⊢ Y iff {¬∀x¬(X → Y ), ∀ xX , ¬Y } is inconsistent. iff ∀ xX , ¬Y ⊢ ∀x¬(X → Y ). ∀ xX P ∀ xX → X A4 X MP X → (¬Y → ¬(X → Y )) Th ¬Y → ¬(X → Y ) MP ¬Y P ¬(X → Y ) MP  ∀x¬(X → Y ) UG

⊢ ∀x(X → Y ) → (¬∀ x¬X → ¬∀ x¬Y ) ⊢ ∀x(X → Y ) → (¬∀ x¬X → ¬∀ x¬Y ) Iff {∀x(X → Y ), (∀ x¬Y ) } ⊢ ∀ x¬X by deduction theorem Iff {∀x(X → Y ), ∀ x¬X , ∀ x¬Y } is inconsistent, by RAA Iff ∀x(X → Y ),(∀ x¬Y ) ⊢ ∀ x¬X , by RAA Proof

⊢ ∀x((x ≈ f(y)) → Qx ) → Qf (y). ∀x((x ≈ f(y)) → Qx ) P ∀x((x ≈ f(y)) → Qx ) → ((f(y) ≈ f(y)) → Qf (y)) A4,   (f(y) ≈ f(y)) → Qf (y) MP  f(y) ≈ f(y) A6 Qf (y) MP ∀x((x ≈ f(y)) → Qx ) ⊢ Qf (y). ⊢ ∀x((x ≈ f(y)) → Qx ) → Qf (y)

{Pa, ∀x( Px → Qx ), ∀x(Rx → ¬ Qx ), Rb } ⊢ ¬(a ≈ b). ∀x( Px → Qx ) P ∀x( Px → Qx ) → (Pa → Qa ) A4 Pa → Qa MP Pa P  1. Qa MP a ≈ b P (a ≈ b) → ( Qa → Qb ) A7 Qa → Qb MP 2. Qb 1,MP ∀x(Rx → ¬ Qx ) P ∀x(Rx → ¬ Qx ) → ( Rb → ¬ Qb ) A4  Rb → ¬ Qb MP Rb P 3. ¬ Qb MP

∀ x∀y (f(x, y) ≈ f(y, x)), ∀ x∀y (f(x, y) ≈ y) ⊨ ¬∀x¬∀y(x ≈ y) 1. ∀ x∀y (f(x, y) ≈ y) P 2. ∀ x∀y (f(x, y) ≈ y) → ∀ y∀x (f(y, x) ≈ x) Th 3. ∀ y∀x (f(y, x) ≈ x) MP 4. f(x, y) ≈ y 1,A4,MP 5. f(y, x) ≈ x 3,A4,MP 6. ∀ x∀y (f(x, y) ≈ f(y, x)) P 7. f(x, y) ≈ f(y, x) A4,MP 8. x ≈ y 4,5,7,A7,MP 9. ∀y(x ≈ y) UG 10. ∀x¬∀y(x ≈ y) P 11. ∀x¬∀y(x ≈ y) → ¬∀y(x ≈ y) A4 12. ¬∀y(x ≈ y) MP

⊢ ¬(∀ xX → Y ) → ∀x¬(X → Y ) iff ¬(∀ xX → Y ) ⊢ ∀x¬(X → Y ) iff {¬(∀ xX → Y ), ¬∀x¬(X → Y )} is inconsistent. iff ¬∀x¬(X → Y ) ⊢ ∀ xX → Y iff {¬∀x¬(X → Y ), ∀ xX } ⊢ Y iff {¬∀x¬(X → Y ), ∀ xX , ¬Y } is inconsistent. iff {∀ xX , ¬Y } ⊢ ∀x¬(X → Y ) ∀ xX P  ∀ xX → X A4 X MP X → (¬Y → ¬(X → Y)) Th ¬Y P (¬Y → ¬(X → Y)) MP ¬(X → Y) MP   ∀x ¬(X → Y) UG If x is not a free variable of Y ⊢ ¬(∀ xX → Y ) → ∀x¬(X → Y )  

Laws in fl First order Logic

LAWS Formulas X, Y, variables x, y, and terms r, s, t Constants: ∀x(⊥ → X) ≡ ⊤, ∃x(⊥ ∧ X) ≡ ⊥. Equality: (t ≈ t) ≡ ⊤, (s ≈ t) ≡ (t ≈ s), {r ≈ s, s ≈ t} ≡ (r ≈ t), {s ≈ t, X[x/s]} ≡ X[x/t].   One-Point Rule: If x does not occur in t, then ∀x((x ≈ t) → X) ≡ X[x/t] and ∃x((x ≈ t) ∧ X) ≡ X[x/t]. Empty Quantification: If x does not occur free in X, then ∀ xX ≡ X and ∃ xX ≡ X. De Morgan: ¬∀ xX ≡ ∃ x¬X , ¬∃ xX ≡ ∀ x¬X , ∀ xX ≡ ¬∃ x¬X , ∃ xX ≡ ¬∀ x¬X .  Renaming: If x does not occur free in X, then ∀ yX ≡ ∀x(X[y/x]) and ∃ yX ≡ ∃x(X[y/x]).

LAWS Commutativity: ∀ x∀yX ≡ ∀ y∀xX , ∃ x∃yX ≡ ∃ y∃xX , ∃ x∀yX _ ∀ y∃xX . Distributivity: ∀x(X ∧ Y ) ≡ ∀ xX ∧ ∀ xY , ∃x(X ∨ Y ) ≡ ∃ xX ∨ ∃ xY , ∀ xX ∨ ∀ xY ≡ ∀x(X ∨ Y ), ∃x(X ∧ Y ) ≡ ∃ xX ∧ ∃ xY . If x does not occur free in X, then ∀x(X ∨ Y ) ≡ X ∨ ∀ xY , ∃x(X ∧ Y ) ≡ X ∧ ∃ xY , ∀x(X → Y ) ≡ X → ∀ xY , ∃x(X → Y ) ≡ X → ∃ xY , ∀x(Y → X) ≡ ∃ xY → X, ∃x(Y → X) ≡ ∀ xY → X.

Four Quantifier LAWS Let x be a variable free for a term t in a formula X. Let α be a parameter not mentioned in Σ ∪ { X} Universal Specifica tion (US): ∀ xX ⊢ X[x/t]. Existential Generalization (EG): X[x/t] ⊢ ∃ xX . Universal Generalization (UG) : If Σ ⊢ X[x/ α], then Σ ⊢ ∀ xX . Existential Specification (ES): If Σ ∪ { X[x/ α]} ⊢ Y , then Σ ∪ {∃ xX } ⊢ Y.

∀ x∀y (f(x, y) ≈ f(y, x)), ∀ x∀y (f(x, y) ≈ y), ∀x¬∀y(x ≈ y) ⊨ ¬∀y(x ≈ y) ∀ x∀y (f(x, y) ≈ y) P 2. ∀ x∀y (f(x, y) ≈ y) → ∀ y∀x (f(y, x) ≈ x) Th 3. ∀ y∀x (f(y, x) ≈ x) MP 4. f(x, y) ≈ y 1,A4,MP 5. f(y, x) ≈ x 3,A4,MP 6. ∀ x∀y (f(x, y) ≈ f(y, x)) P 7. f(x, y) ≈ f(y, x) A4,MP 8. x ≈ y 4,5,7,A7,MP 9. ∀y(x ≈ y) UG 10. ∀x¬∀y(x ≈ y) P 11. ∀x¬∀y(x ≈ y) → ¬∀y(x ≈ y) A4 12. ¬∀y(x ≈ y) MP

Natural deduction

Inference rules where c is a new constant not occurring in Y . where y is a new variable

{∀ x ( P xy → Q x ), ∀ z P zy } ⊢ ∀ x Q x 1. ∀x( Pxy → Qx ) P 2. ∀ zPzy P 3. Puy → Qu ∀e, [x/u] 4. Puy  2, ∀e 6. ∀ xQx ∀ i u 5. Qu →e Scope of new variable (u)

∀ x ( P xy → Q x ), ∃ z P zy ⊢ ∃ x Q x 1. ∀x( Pxy → Qx ) P 2. ∃ zPzy P 3. Pcy 2, ∃e 4. Pcy → Qc 1, ∀e 5. Qc →e 6. ∃ xQx ∃ i Scope of new variable (c) c

exercises ⊢ ∀ xX → X[x/t] for any term t free for x in X. ⊢ ∀ x(X → Y ) → (X → ∀ xY ) if x is not free in X. ⊢ (s ≈ t) → (X[x/s] → X[x/t]). Pa, ∀x( Px → Qx ), ∀x(Rx → ¬ Qx ), Rb |= ¬(a ≈ b). ∀x(Lx → Fx ) ⊢ ∀x(∃y(Ly ∧ Sxy ) → ∃y( Fy ∧ Sxy ))

1. ⊢ ∀ xX → X[x/t] for any term t free for x in X X[x/t] ∀e ∀ xX CP ∀ xX → X[x/t] → i

2. ⊢ ∀ x(X → Y ) → (X → ∀ xY ) if x is not free in X. 1. ∀x(X → Y ) CP X CP y X → Y [x/y] 1, ∀e Y [x/y] → e ∀ xY ∀ i X → ∀ xY → i ∀x(X → Y ) → (X → ∀ xY ) → i

3. ⊢ (s ≈ t) → (X[x/s] → X[x/t]). (s ≈ t) CP X[x/s] CP X[x/t] ≈e X[x/s] → X[x/t] → i (s ≈ t) → (X[x/s] → X[x/t]) → i

4. Pa, ∀x(Px → Qx ), ∀x(Rx → ¬ Qx ), Rb ⊨ ¬(a ≈ b) 3. Pa P 1. ∀x( Px → Qx ) P 6. ∀x(Rx → ¬ Qx ) P 5. Rb P 2. Pa → Qa ∀e 4. Qa →e 7. Rb → ¬ Qb ) ∀e 8. ¬ Qb 5,7 → e 9. a ≈ b CP(assume) 10. ¬ Qa 8,9, ≈ 11. ⊥ 4,10 12. ¬ ( a ≈ b) ¬ i

5. ∀x(Lx → Fx ) ⊢ ∀x(∃y(Ly ∧ Sxy ) → ∃y( Fy ∧ Sxy )) 1. ∀x(Lx → Fx ) P 2. ∃y(Ly ∧ Sxy ) CP 3. Lc ∧ Sxc c 4. Lc ∧e 5. Lc → Fc 1, ∀e 6. Fc →e 7. Sxc 3, ∧e 8. Fc ∧ Sxc ∧ i 9. ∃y( Fy ∧ Sxy )) ∃ i 10. ∃y( Fy ∧ Sxy )) ∃e 11. ∃y(Ly ∧ Sxy ) → ∃y( Fy ∧ Sxy )) → i

∀x(Lx → Fx ) ⊢ ∀x(∃y(Ly ∧ Sxy ) → ∃y( Fy ∧ Sxy ))

Analytic tableaux

Propositional Tableau Tree whose root is the proposition P and generated by applying PT-rules. Children of a node is the denominator of the corresponding rule Path: From root to a leaf Complete Path : Rule applied on every compound proposition Closed Path: Contains ⊥ or p and ¬p for some atomic or constant proposition Open Path: Path which is not closed Stacking rules Branching rules Completed Closed tableau  Open tableau

Example (p ∨ q )∧( q ∨ r) 1. p ∨ q 2. q ∨ r 3. p q 4. q r q r

Show {p→(¬ q→r ),p→¬q,¬( p→r )} inconsistent 1. p→(¬ q→r ) 2. p→¬q 3. ¬( p→r ) 4. p 5. ¬r 6. ¬p 7. ¬q 8. ¬p. 9. ¬ q→r . 10. ¬p. 11. ¬ q→r 12. ¬¬q 13. r 14. ¬¬q 15. r 16. q 17. q x x x x x x Closed Tableau hence Inconsistent

Show {p→(¬ q→r ),p→¬q,¬( p→r )} inconsistent 1. p→(¬ q→r ) 2. p→¬q 3. ¬( p→r ) 4. p 5. ¬r 6. ¬p 7. ¬q 8. ¬p 9. ¬ q→r 10. ¬¬q 11. r 12. q x x x x

Additional rules of inference The restriction of ‘t is a new term’ means that if s is a sub-term of t, then s neither occurs in The current formula, Nor in any premise used so far in the path, Nor in any formula introduced to the path by an existential rule. Apply all stacking rules before applying any branching rule whenever possible Apply existential rules before applying universal rules. t is a constant that has not occurred earlier in the path. Use the (∃) and (¬∀) rules before applying the (∀) and (¬∃) rules. 

⊢ ∀ x X → X[ x/t ] where x is free for term t in X. ¬(∀ xX → X[x/t])  1. ∀ xX ¬X[x/t] X[x/t] 1, (∀)  ¬ (¬∀ xX [x/t]) )   (∀ xX [x/t]) )   ∀ xX X

⊢ ∀ x (X → Y ) → (X → ∀ x Y ) where x is not free in X ¬(∀x(X → Y ) → (X → ∀ xY )   1. ∀x(X → Y ) ¬(X → ∀ xY ) X ¬ ∀ xY ¬Y [x/c ] new constant c X → Y [x/c] 1, x not free in X ¬X Y [x/c]  X → Y = ¬X ∨ Y

∃ x∃y∃z (¬ Qx ∧¬ Qy ∧¬ Qz ∧ Qf (f(x, y), z)) ⊢ ∃ x∃y (¬ Qx ∧¬ Qy ∧ Qf (x, y)).

exercises ⊢ (t ≈ t) ⊢ (s ≈ t) → (X[x/s] → X[x/t]), where x is free for terms s, t in X. ⊢ ∃x(∃ yPy → Px ) {∃ xPx , ¬Pc}. {∃ xPx ∧ ∃ xQx , ¬∃x( Px ∧ Qx ), ∀ xPx → Ps}.  {∀x( Px → Qx ), ∃ xPx , ∀ x¬Qx , ∃ xPx ∨ ¬Pc}. ∀x(∃y( Pxy ∧ Qy ) → ∃z( Rz ∧ Pxz )) ⊨ ¬ ∃ xRx → ∀ x∀y ( Pxy → ¬ Qy ). {∀ x∃yPxy , ∀ x∀y∀z (( Pxy ∧ Pyz ) → Pxz ), ¬∃ xPxx }.

Solutions ⊢ (t ≈ t) ¬(t ≈ t) (t ≈ t) (≈ ) X

Additional exercises {∃ xPx , ¬Pc}. {∃ xPx ∧ ∃ xQx , ¬∃x( Px ∧ Qx ), ∀ xPx → Ps}.  {∀x( Px → Qx ), ∃ xPx , ∀ x¬Qx , ∃ xPx ∨ ¬Pc}. ∀x(∃y( Pxy ∧ Qy ) → ∃z( Rz ∧ Pxz )) _ ¬∃ xRx → ∀ x∀y ( Pxy → ¬ Qy ). {∀ x∃yPxy , ∀ x∀y∀z (( Pxy ∧ Pyz ) → Pxz ), ¬∃ xPxx }.
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