Example: Proving a Congruence

Temporal Verification of Reactive Systems: Safety

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Unformatted text preview: CS256/Winter 2007 — Lecture #4 Zohar Manna Announcements • HW1 solutions and grades available. 4-1 Example: Proving a Congruence For temporal formulas ϕ and ψ , show 1 0 ϕ ∧ 1 0 ψ ≈ 1 ( ϕ ∧ ψ ) We have to show 1 0 ϕ ∧ 1 0 q ⇒ 1 ( ϕ ∧ ψ and 1 ( ϕ ∧ ψ ) ⇒ 1 0 ϕ ∧ 1 0 ψ ⇒ The left-to-right entailment is valid: Consider arbitrary σ and j such that ( σ, j ) q 1 0 ϕ ∧ 1 0 ψ. Thus ∃ k 1 ≥ j. ( σ, k 1 ) q 0 ϕ and ∃ k 2 ≥ j. ( σ, k 2 ) q 0 ψ 4-2 Example: Proving a Congruence (Cont’d) 1 0 ϕ ∧ 1 0 ψ ≈ 1 ( ϕ ∧ ψ ) Unraveling the definition of , we get ∃ k 1 ≥ j. ∀ k 1 ≥ k 1 . ( σ, k 1 ) q ϕ and ∃ k 2 ≥ j. ∀ k 2 ≥ k 2 . ( σ, k 2 ) q ψ. This implies that k = max { k 1 ,k 2 } z }| { ∃ k ≥ j. ∀ k ≥ k. ( σ, k ) q ϕ and ( σ, k ) q ψ. So ∃ k ≥ j. ( σ, k ) q ( ϕ ∧ ψ ) . That is, ( σ, j ) q 1 ( ϕ ∧ ψ ) . ⇐ The right-to-left entailment is valid. All implications in the first part hold in reverse, so the entailment is valid. 4-3 Example: Proving an Equivalence / Disproving a Congruence For temporal logic formulas ϕ and ψ , show 1 ϕ ∼ 1 Q ϕ 1 ϕ 6≈ 1 Q ϕ We shall prove: (1) 1 ϕ ⇒ 1 Q ϕ is valid; Thus ϕ → 1 Q ϕ is valid. (2) 1 Q ϕ → 1 ϕ is valid. (3) 1 Q ϕ ⇒ 1 ϕ is not valid. 4-4 (1) 1 ϕ ⇒ 1 Q ϕ is valid: Consider arbitrary σ and j such that ( σ, j ) q 1 ϕ . Then ∃ i ≥ j. ( σ, i ) q ϕ . Hence ∃ i ≥ j. ∃ k : 0 ≤ k ≤ i. | {z } k = i ( σ, k ) q ϕ . By def. ∃ i ≥ j. ( σ, i ) q Q ϕ . Therefore ( σ, j ) q 1 Q ϕ . (2) 1 Q ϕ → 1 ϕ is valid: Consider arbitrary σ such that ( σ, 0) q 1 Q ϕ . Then ∃ i ≥ . ( σ, i ) q Q ϕ . Hence ∃ i ≥ . ∃ k : 0 ≤ k ≤ i. ( σ, k ) q ϕ . Hence ∃ k ≥ . ( σ, k ) q ϕ . Therefore ( σ, 0) q 1 ϕ . (3) 1 Q ϕ ⇒ 1 ϕ is not valid. Counterexample: Take ϕ : p (propositional symbol) σ = h s : p, s 1 : ¬ p, s 2 : ¬ p, s 3 : ¬ p, . . . i and j = 1 Then ( σ, 1) q 1 Q p , but ( σ, 1) q / 1 p ....
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Example: Proving a Congruence - CS256/Winter 2007 —...

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