SUPERPOSITION FOR LAMBDA-FREE HIGHER-ORDER LOGIC Motivation: - - PowerPoint PPT Presentation

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SUPERPOSITION FOR LAMBDA-FREE HIGHER-ORDER LOGIC Motivation: - - PowerPoint PPT Presentation

Oct 11, 2022 119 likes •304 views

1 ALEXANDER BENTKAMP JASMIN BLANCHETTE SIMON CRUANES UWE WALDMANN SUPERPOSITION FOR LAMBDA-FREE HIGHER-ORDER LOGIC Motivation: Sledgehammer 2 Proof goal from Isabelle Fact selection Translation to FOL A complete HO

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SUPERPOSITION FOR 
 LAMBDA-FREE HIGHER-ORDER LOGIC

ALEXANDER BENTKAMP JASMIN BLANCHETTE SIMON CRUANES UWE WALDMANN

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SLIDE 2

Motivation: Sledgehammer

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Fact selection Translation to FOL Superposition provers Proof reconstruction Proof goal 
 from Isabelle Proof text
 in Isabelle LEO-II/III Satallax A complete HO superposition prover SMT provers

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DESIGN PRINCIPLE: BE GRACEFUL

HO superposition on first-order problems should coincide with FO superposition

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FOL λ-free HOL / applicative FOL predicate-free HOL HOL

Our way to higher-order superposition

partial application 
 & applied variables λ-expressions /
 comprehension 
 axioms boolean formulas 
 nested in terms

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Translation to FOL: applicative encoding

NOT GRACEFUL!

f (H f) app(f, app(H, f)) is translated to λ-free HOL FOL

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Term orders for λ-free HOL

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Compatibility with arguments?
 t > s ⇒ t u > s u No:
 This talk

LPO KBO with argument 
 coefficients

Yes:
 Petar’s talk

KBO without argument 
 coefficients

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SLIDE 7

The superposition rule

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C ∨ (¬) s[u] = s’ (D ∨ C ∨ (¬) s[t’] = s’)σ σ = mgu(t,u) D ∨ t = t’

+ order conditions

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Superposition only at argument subterms

f a (h b c) f a (h b c)

Prefix subterms: Argument subterms:

g = f g a ≠ b f a ≠ b SUP

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Argument congruence rule C ∨ t = s C ∨ t X = s X ARGCONG g = f g a ≠ b f a ≠ b SUP g X = f X ARGCONG

Example:

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Argument congruence rule C ∨ t = s C ∨ t X = s X ARGCONG

BUT ISN’T THIS RULE ALWAYS REDUNDANT?

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Floor encoding

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Encode ground λ-free HOL terms into FOL:

⎣f a⎦= f1(a0) ⎣f⎦= f0

Redundancy is defined with respect to this encoding.

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Floor encoding

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Example: g = f g X = f X g0 = f0 g1 a0 = f1 a0 ARGCONG

Not redundant!

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What changes in the proof?

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Refutational completeness: 
 Let N be saturated up to redundancy, ⊥∉ N. 
 Then N has a model.

N G(N) model of G(N) model of N

model construction

Proof sketch for FOL:

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What changes in the proof?

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Refutational completeness: 
 Let N be saturated up to redundancy, ⊥∉ N. 
 Then N has a model.

N G(N) model of G(N) model of N ⎣G(N)⎦ model of⎣G(N)⎦

model construction

Proof sketch for λ-free HOL:

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Issue: superposition at variables

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C = … X … X a …

Given g > f, it is unclear whether X := g or X:= f 
 will yield the smaller clause Solution #1: 
 purifying calculus … X u̅ … X v̅ … … X u̅ … Y v̅ … ∨ X ≠ Y

is purified to if u̅ ≠ v̅

Solution #2: 
 nonpurifying calculus Perform superpositions at variables 
 if the order situation is unclear

Example:

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Evaluation of our prototype

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TPTP benchmarks Judgment Day 
 λ-free HOL benchmarks

# unsat

FO HO 32 facts 512 facts

first-order mode

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  • applicative encoding

151 677 873 843

purifying calculus

180 647 851 908

nonpurifying calculus

179 669 866 889

using the Zipperposition theorem prover

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In summary

  • We developed refutationally complete calculi 


for λ-free HOL


  • They reduce the gap between HO proof assistants 


and superposition provers


  • They are promising as a stepping stone towards a 


HO superposition calculus

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