Lecture 13 Resolution-Based Inference 11 th February 2020 - - PowerPoint PPT Presentation

Informatics 2D ⋅ Agents and Reasoning ⋅ 2019/2020

Lecture 13 ⋅ Resolution-Based Inference

Claudia Chirita

School of Informatics, University of Edinburgh

11th February 2020

Based on slides by: Jacques Fleuriot, Michael Rovatsos, Michael Herrmann, Vaishak Belle

Previously on INF2D

Backward chaining – if Goal is known (goal directed) – can query for data Forward chaining – if specifjc Goal is not known, but the system needs to react to new facts (data driven) – can make suggestions What do users expect from the system? Which direction has the larger branching factor?

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Limitations

...due to restriction to defjnite clauses In order to apply GMP

• premises of rules contain only non-negated symbols
• the conclusion of any rule is a non-negated symbol
• facts are non-negated atomic sentences

Possible solution: introduce more variables, e.g. 𝑅 ∶= ¬𝑄 What about: “If we cannot prove 𝐵, then ¬𝐵 is true”? (works only if there is a rule for each variable)

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Resolution one more time

• Negate query 𝛽.
• Convert everything to CNF.
• Repeat: Choose clauses and resolve (based on unifjcation).
• If resolution results in empty clause, 𝛽 is proved.
• Return all substitutions (or Fail).

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Ground Binary Resolution & Modus Ponens

Ground binary resolution

𝐷 ∨ 𝑄 𝐸 ∨ ¬𝑄 𝐷 ∨ 𝐸

Suppose 𝐷 = False.

𝑄 ¬𝑄 ∨ 𝐸 𝐸

i.e. 𝑄 and 𝑄 → 𝐸 entails 𝐸. Modus ponens is a special case of binary resolution.

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Full Resolution & Generalised Modus Ponens

GMP with 𝑞

𝜄 = 𝑞𝜄

𝑞

, 𝑞 , … , 𝑞

• (𝑞 ∧ 𝑞 ∧ … ∧ 𝑞 → 𝑟)

𝑟𝜄 𝑞

, 𝑞 , … , 𝑞

• (𝑟 ∨ ¬𝑞 ∨ ¬𝑞 ∨ … ∨ ¬𝑞)

𝑟𝜄

Full resolution with 𝜄 mgu of all 𝑄 and 𝑄

• 𝐷 ∨ 𝑄 ∨ … ∨ 𝑄

𝐸 ∨ ¬𝑄

∨ … ∨ ¬𝑄

• (𝐷 ∨ 𝐸) 𝜄

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Resolution in Implication Form

Ground binary resolution

𝐷 ∨ 𝑄 𝐸 ∨ ¬𝑄 𝐷 ∨ 𝐸

Set 𝐷 = ¬𝐵.

𝐵 → 𝑄 𝑄 → 𝐸 𝐵 → 𝐸

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Example ⋅ Memes and Theorems

• Some students like all memes.

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.𝑁(𝑧) → Likes(𝑦, 𝑧)

• No student likes any theorem.

𝐺 ∶ ∀𝑦, 𝑧.𝑇(𝑦) ∧ 𝑈(𝑧) → ¬Likes(𝑦, 𝑧)

• Show: No meme is a theorem.

𝐺 ∶ ∀𝑦.𝑁(𝑦) → ¬𝑈(𝑦)

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Example ⋅ Memes and Theorems

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Example ⋅ Memes and Theorems

CNF ⋅ Eliminating implications

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.𝑁(𝑧) → Likes(𝑦, 𝑧) ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.¬𝑁(𝑧) ∨ Likes(𝑦, 𝑧) 𝐺 ∶ ∀𝑦, 𝑧.𝑇(𝑦) ∧ 𝑈(𝑧) → ¬Likes(𝑦, 𝑧) ∀𝑦, 𝑧.¬𝑇(𝑦) ∨ ¬𝑈(𝑧) ∨ ¬Likes(𝑦, 𝑧) 𝐺 ∶ ∀𝑦.𝑁(𝑦) → ¬𝑈(𝑦) ∀𝑦.¬𝑁(𝑦) ∨ ¬𝑈(𝑦)

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Example ⋅ Memes and Theorems

CNF ⋅ Standardising variables apart, skolemising, dropping universal quantifjers

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.¬𝑁(𝑧) ∨ Likes(𝑦, 𝑧) 𝑇(𝐻) ∧ (¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧)) 𝐺 ∶ ∀𝑦, 𝑧.¬𝑇(𝑦) ∨ ¬𝑈(𝑧) ∨ ¬Likes(𝑦, 𝑧) ¬𝑇(𝑥) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝑥, 𝑨) 𝐺 ∶ ∀𝑦.¬𝑁(𝑦) ∨ ¬𝑈(𝑦) ¬𝑁(𝑦) ∨ ¬𝑈(𝑦)

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Example ⋅ Memes and Theorems

Unifjcation

𝐺 ∶ 𝑇(𝐻) ∧ (¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧)) 𝐺 ∶ ¬𝑇(𝑥) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝑥, 𝑨) 𝑥/𝐻 ∶ ¬𝑇(𝐻) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨)

Negation of proof goal

¬(¬𝑁(𝑦) ∨ ¬𝑈(𝑦)) ≡ 𝑁(𝑦) ∧ 𝑈(𝑦)

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Example ⋅ Memes and Theorems

𝑇(𝐻) ∧ (¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧)) ¬𝑇(𝐻) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨) 𝑁(𝑦) ∧ 𝑈(𝑦)

Clauses: 𝑇(𝐻), 𝑁(𝑦), 𝑈(𝑦), ¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧),

¬𝑇(𝐻) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨) 𝑇(𝐻) ¬𝑇(𝐻) ∨ ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨) ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨) ¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧) ¬𝑈(𝑨) ∨ ¬Likes(𝐻, 𝑨) ¬𝑁(𝑨) ∨ ¬𝑈(𝑨)

Substitute 𝑨/𝑦

¬𝑁(𝑦) ∨ ¬𝑈(𝑦) 𝑁(𝑦) ¬𝑈(𝑦)

and

¬𝑈(𝑦) 𝑈(𝑦)

• Therefore, ¬𝑁(𝑦) ∨ ¬𝑈(𝑦), i.e. 𝑁(𝑦) → ¬𝑈(𝑦).

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Example ⋅ Memes and Theorems 2.0

• Some students like all memes.

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.𝑁(𝑧) → Likes(𝑦, 𝑧)

• No student likes any theorem.

𝐺 ∶ ∀𝑦.𝑇(𝑦) → ∀𝑧.𝑈(𝑧) → ¬Likes(𝑦, 𝑧)

• Show: No meme is a theorem.

𝐺 ∶ ∀𝑦.𝑁(𝑦) → ¬𝑈(𝑦)

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Example ⋅ Memes and Theorems 2.0

CNF ⋅ Eliminating implications

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.𝑁(𝑧) → Likes(𝑦, 𝑧) ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.¬𝑁(𝑧) ∨ Likes(𝑦, 𝑧) 𝐺 ∶ ∀𝑦.𝑇(𝑦) → ∀𝑧.𝑈(𝑧) → ¬Likes(𝑦, 𝑧) ∀𝑦.¬𝑇(𝑦) ∨ ∀𝑧.¬𝑈(𝑧) ∨ ¬Likes(𝑦, 𝑧) 𝐺 ∶ ∀𝑦.𝑁(𝑦) → ¬𝑈(𝑦) ∀𝑦.¬𝑁(𝑦) ∨ ¬𝑈(𝑦)

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Example ⋅ Memes and Theorems 2.0

CNF ⋅ Standardising variables apart, skolemising, dropping universal quantifjers

𝐺 ∶ ∃𝑦.𝑇(𝑦) ∧ ∀𝑧.¬𝑁(𝑧) ∨ Likes(𝑦, 𝑧) 𝑇(𝐻) ∧ (¬𝑁(𝑧) ∨ Likes(𝐻, 𝑧)) 𝐺 ∶ ∀𝑦.¬𝑇(𝑦) ∨ ∀𝑧.¬𝑈(𝑧) ∨ ¬Likes(𝑦, 𝑧) ¬𝑇(𝑥) ∨ (¬𝑈(𝑨) ∨ ¬Likes(𝑥, 𝑨)) 𝐺 ∶ ∀𝑦.¬𝑁(𝑦) ∨ ¬𝑈(𝑦) ¬𝑁(𝑦) ∨ ¬𝑈(𝑦)

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Resolution ⋅ Soundness and completeness

Resolution is sound and complete. A set of clauses 𝑇 is unsatisfjable if and only if one can derive the empty clause (false) from 𝑇. Soundness: derivability of empty clause implies unsatisfjability. Can be proved by noticing that every model that satisfjes the premises of resolution satisfjes also satisfjes its conclusion. Completeness: every unsatisfjable clause can be refuted by resolution. Can be proved using completeness of propositional resolution and lifting (as in the following slides; the full proof is beyond the scope of this course).

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Resolution ⋅ Completeness proof

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Completeness proof ⋅ Step 1

For a set of clauses 𝑇, we call the Herbrand universe of 𝑇 the set 𝐼 of all ground terms that can be constructed from the function symbols in 𝑇. Example For 𝑇 = {¬𝑄(𝑦, 𝐺(𝑦, 𝐵)) ∨ ¬𝑅(𝑦, 𝐵) ∨ 𝑆(𝑦, 𝐶)} we have

𝐼 = {𝐵, 𝐶, 𝐺(𝐵, 𝐵), 𝐺(𝐵, 𝐶), 𝐺(𝐶, 𝐵), 𝐺(𝐶, 𝐶), 𝐺(𝐵, 𝐺(𝐵, 𝐵)), …}

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Completeness proof ⋅ Step 1

For a set of clauses 𝑇 and 𝑄 a set of ground terms,

𝑄(𝑇), the saturation of 𝑇 with respect to 𝑄, is the set of all

ground clauses obtained by applying all possible consistent substitutions of variables in 𝑇 with ground terms from 𝑄. The saturation of a set 𝑇 with respect to its Herbrand universe is called the Herbrand base of 𝑇 and denoted 𝐼(𝑇). Example

𝐼(𝑇) = {¬𝑄(𝐵, 𝐺(𝐵, 𝐵)) ∨ ¬𝑅(𝐵, 𝐵) ∨ 𝑆(𝐵, 𝐶), ¬𝑄(𝐶, 𝐺(𝐶, 𝐵)) ∨ ¬𝑅(𝐶, 𝐵) ∨ 𝑆(𝐶, 𝐶), ¬𝑄(𝐺(𝐵, 𝐵), 𝐺(𝐺(𝐵, 𝐵), 𝐵)) ∨ ¬𝑅(𝐺(𝐵, 𝐵), 𝐵) ∨ 𝑆(𝐺(𝐵, 𝐵), 𝐶), ¬𝑄(𝐺(𝐵, 𝐶), 𝐺(𝐺(𝐵, 𝐶), 𝐵)) ∨ ¬𝑅(𝐺(𝐵, 𝐶), 𝐵) ∨ 𝑆(𝐺(𝐵, 𝐶), 𝐶) , … }

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Completeness proof ⋅ Step 1

Herbrand’s theorem (1930) If a set 𝑇 of clauses is unsatisfjable, then there exists a fjnite subset of 𝐼(𝑇) that is also unsatisfjable.

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Completeness proof ⋅ Step 2

Let 𝑇 be that fjnite unsatisfjable subset of ground sentences. Running propositional resolution to completion on 𝑇 will derive a contradiction.

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Completeness proof ⋅ Step 3

Lifting lemma Let 𝐷 and 𝐷 be two clauses with no shared variables, and let 𝐷

and 𝐷 ground instances of 𝐷 and 𝐷.

If 𝐷 is a resolvent of 𝐷

and 𝐷 , then there exists a clause 𝐷

such that:

𝐷 is a resolvent of 𝐷 and 𝐷 𝐷 is a ground instance of 𝐷.

𝐷, 𝐷

instantiation

𝐷

, 𝐷

• ↓

↓

• lifting
• ↓

↓ 𝐷

instantiation

𝐷

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Completeness proof ⋅ Step 3

Example

𝐷 = ¬𝑄(𝑦, 𝐺(𝑦, 𝐵)) ∨ ¬𝑅(𝑦, 𝐵) ∨ 𝑆(𝑦, 𝐶) 𝐷 = ¬𝑂(𝐻(𝑧), 𝑨) ∨ 𝑄(𝐼(𝑧), 𝑨) 𝐷

= ¬𝑄(𝐼(𝐶), 𝐺(𝐼(𝐶), 𝐵)) ∨ ¬𝑅(𝐼(𝐶), 𝐵) ∨ 𝑆(𝐼(𝐶), 𝐶)

𝐷

= ¬𝑂(𝐻(𝐶), 𝐺(𝐼(𝐶), 𝐵)) ∨ 𝑄(𝐼(𝐶), 𝐺(𝐼(𝐶), 𝐵))

𝐷 = ¬𝑂(𝐻(𝐶), 𝐺(𝐼(𝐶), 𝐵)) ∨ ¬𝑅(𝐼(𝐶), 𝐵) ∨ 𝑆(𝐼(𝐶), 𝐶) 𝐷 = ¬𝑂(𝐻(𝑧), 𝐺(𝐼(𝑧), 𝐵)) ∨ ¬𝑅(𝐼(𝑧), 𝐵) ∨ 𝑆(𝐼(𝑧), 𝐶)

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Effjcient algorithms for resolution

Heuristics to make resolution more effjcient: Unit preference: prefer clauses with only one symbol. Pure clauses: a pure clause contains symbol 𝐵 which does not occur in any other clause. Cannot lead to contradiction. Tautology: clauses containing 𝐵 and ¬𝐵. Set of support: identify useful clauses and ignore the rest. Input resolution: intermediately generated clauses can only be combined with original input clauses. Subsumption: if a clause contains another one, use only the shorter clause. Prune unnecessary facts from the KB. Including heuristics, resolution is more effjcient than DPLL.

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Summary

• Limitations of GMP
• Relationship between inference rules
• Completeness of resolution – the lifting lemma
• Effjcient algorithms for resolution

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