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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

  1. Theorem Prover Moca (Oi and Hirokawa, JAIST) � ∗ � � Infeasibility Problem : s i σ � − → R t i σ for every σ ? i

  2. Theorem Prover Moca (Oi and Hirokawa, JAIST) � ∗ � � Infeasibility Problem : s i σ � − → R t i σ for every σ ? i transformation Horn Satisfiability Problem : M | = Φ for some M ?

  3. Theorem Prover Moca (Oi and Hirokawa, JAIST) � ∗ � � Infeasibility Problem : s i σ � − → R t i σ for every σ ? i transformation Horn Satisfiability Problem : M | = Φ for some M ? transformation Word Problem : T �≈ E F?

  4. Theorem Prover Moca (Oi and Hirokawa, JAIST) � ∗ � � Infeasibility Problem : s i σ � − → R t i σ for every σ ? i transformation Horn Satisfiability Problem : M | = Φ for some M ? transformation Word Problem : T �≈ E F? Maximal Ordered Completion with Approximation YES / MAYBE 1/3

  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 )

  6. Transformations to Word Problem Infeasibility Problem Horn Satisfiability Problem CTRS: Formulas: x − 0 → x ∀ x. x − 0 = x 0 − x → 0 ∀ x. 0 − x = 0 s ( x ) − s ( y ) → x − y ∀ x, y. s ( x ) − s ( y ) = x − y ∀ x. x − x � = s ( x ) Condition: ∗ x − x − → R s ( x )

  7. Transformations to Word Problem Infeasibility Problem Horn Satisfiability Problem Word Problem ES: CTRS: Formulas: x − 0 = x x − 0 → x ∀ x. x − 0 = x 0 − x = 0 0 − x → 0 ∀ x. 0 − x = 0 s ( x ) − s ( y ) = x − y s ( x ) − s ( y ) → x − y ∀ x, y. s ( x ) − s ( y ) = x − y f ( x − x, x ) = T ∀ x. x − x � = s ( x ) Condition: f ( s ( x ) , x ) = F ∗ x − x → R s ( x ) − Goal: T = F 2/3

  8. Transformations to Word Problem Infeasibility Problem Horn Satisfiability Problem Word Problem ES: CTRS: Formulas: x − 0 = x x − 0 → x ∀ x. x − 0 = x 0 − x = 0 0 − x → 0 ∀ x. 0 − x = 0 s ( x ) − s ( y ) = x − y s ( x ) − s ( y ) → x − y ∀ x, y. s ( x ) − s ( y ) = x − y f ( x − x, x ) = T ∀ x. x − x � = s ( x ) Condition: f ( s ( x ) , x ) = F ∗ x − x → R s ( x ) − Goal: T = F Infeasible ⇐ = Satisfiable ⇐ = T �≈ E F 2/3

  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

  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

  11. Approximation for Showing T �≈ E F Fact if ≈ E ⊆ ≈ E ′ ( E is approximated as E ′ ) then T �≈ E ′ F implies T �≈ E F Approximation E ′ Original System E x − 0 = x x − 0 = x 0 − x = 0 0 − x = 0 s ( x ) − s ( y ) = x − y s ( x ) − s ( y ) = x − y f ( x − x, x ) = T f ( x − x, y ) = T f ( s ( x ) , x ) = F f ( s ( x ) , x ) = F admits no complete TRS admits complete TRS! 3/3

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