Transformations to Word Problem Infeasibility Problem CTRS: x 0 x - - PowerPoint PPT Presentation

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Jul 24, 2023 205 likes •326 views

Theorem Prover Moca (Oi and Hirokawa, JAIST) Infeasibility Problem : s i R t i for every ? i Theorem Prover Moca (Oi and Hirokawa, JAIST) Infeasibility Problem : s i R t

SLIDE 1

Theorem Prover Moca (Oi and Hirokawa, JAIST)

Infeasibility Problem:

i
siσ

∗

− →R tiσ

for every σ?

SLIDE 2

Theorem Prover Moca (Oi and Hirokawa, JAIST)

Infeasibility Problem:

i
siσ

∗

− →R tiσ

for every σ?

Horn Satisfiability Problem: M | = Φ for some M?

transformation

SLIDE 3

Theorem Prover Moca (Oi and Hirokawa, JAIST)

Infeasibility Problem:

i
siσ

∗

− →R tiσ

for every σ?

Horn Satisfiability Problem: M | = Φ for some M?

transformation

Word Problem: T ≈E F?

transformation

SLIDE 4

Theorem Prover Moca (Oi and Hirokawa, JAIST)

Infeasibility Problem:

i
siσ

∗

− →R tiσ

for every σ?

Horn Satisfiability Problem: M | = Φ for some M?

transformation

Word Problem: T ≈E F?

transformation

YES / MAYBE

Maximal Ordered Completion with Approximation

1/3

SLIDE 5

Transformations to Word Problem

Infeasibility Problem

CTRS: x − 0 → x 0 − x → 0 s(x) − s(y) → x − y Condition: x − x

∗

− →R s(x)

SLIDE 6

Transformations to Word Problem

Infeasibility Problem

CTRS: x − 0 → x 0 − x → 0 s(x) − s(y) → x − y Condition: x − x

∗

− →R s(x)

Horn Satisfiability Problem

Formulas: ∀x. x − 0 = x ∀x. 0 − x = 0 ∀x, y. s(x) − s(y) = x − y ∀x. x − x = s(x)

SLIDE 7

Transformations to Word Problem

Infeasibility Problem

CTRS: x − 0 → x 0 − x → 0 s(x) − s(y) → x − y Condition: x − x

∗

− →R s(x)

Horn Satisfiability Problem

Formulas: ∀x. x − 0 = x ∀x. 0 − x = 0 ∀x, y. s(x) − s(y) = x − y ∀x. x − x = s(x)

Word Problem

ES: x − 0 = x 0 − x = 0 s(x) − s(y) = x − y f(x − x, x) = T f(s(x), x) = F Goal: T = F

2/3

SLIDE 8

Transformations to Word Problem

Infeasibility Problem

CTRS: x − 0 → x 0 − x → 0 s(x) − s(y) → x − y Condition: x − x

∗

− →R s(x)

Horn Satisfiability Problem

Formulas: ∀x. x − 0 = x ∀x. 0 − x = 0 ∀x, y. s(x) − s(y) = x − y ∀x. x − x = s(x)

Word Problem

ES: x − 0 = x 0 − x = 0 s(x) − s(y) = x − y f(x − x, x) = T f(s(x), x) = F Goal: T = F Infeasible ⇐ = Satisfiable ⇐ = T ≈E F

2/3

SLIDE 9

Approximation for Showing T ≈E F

Fact

if ≈E ⊆ ≈E′ (E is approximated as E′) then T ≈E′ F implies T ≈E F

3/3

SLIDE 10

Approximation for Showing T ≈E F

Fact

if ≈E ⊆ ≈E′ (E is approximated as E′) then T ≈E′ F implies T ≈E F

Original System E

x − 0 = x 0 − x = 0 s(x) − s(y) = x − y f(x − x, x) = T f(s(x), x) = F admits no complete TRS

3/3

SLIDE 11

Approximation for Showing T ≈E F

Fact

if ≈E ⊆ ≈E′ (E is approximated as E′) then T ≈E′ F implies T ≈E F

Original System E

x − 0 = x 0 − x = 0 s(x) − s(y) = x − y f(x − x, x) = T f(s(x), x) = F admits no complete TRS

Approximation E′

x − 0 = x 0 − x = 0 s(x) − s(y) = x − y f(x − x, y) = T f(s(x), x) = F admits complete TRS!

3/3