ESSLLI 2018 course Logics for Epistemic and Strategic Reasoning in - PowerPoint PPT Presentation

ESSLLI 2018 course Logics for Epistemic and Strategic Reasoning in Multi-Agent Systems Supplementary slides on Operators on binary relations Valentin Goranko Stockholm University ESSLLI 2018 August 6-10, 2018 Sofia University, Bulgaria V Goranko 1 of 21

Auxiliary slides: some operations on binary relations Let W be a set of possible worlds and R a ⊆ W 2 , R b ⊆ W 2 be two binary (accessibility) relations in W . Thus, each R a , R b is a set of ordered pairs in W . Then we can define union, intersection, and composition of R a and R b as the following binary relations in W . • R a ∪ b := R a ∪ R b = { ( u , w ) | uR a w or uR b w } . • R a ∩ b := R a ∩ R b = { ( u , w ) | uR a w and uR b w } . • R a ◦ b := R a ◦ R b = { ( u , w ) | uR a v and vR b w for some v ∈ W } . Examples follow. V Goranko 2 of 21

Operations on binary relations: union b a b a s 1 s 2 a a b b a a s 3 s 4 b V Goranko 3 of 21

Operations on binary relations: union R a ∪ b = R a ∪ R b s 1 s 2 s 3 s 4 V Goranko 4 of 21

Operations on binary relations: intersection b a b a s 1 s 2 b a b a b a s 3 s 4 a b V Goranko 5 of 21

Operations on binary relations: intersection R a ∩ R b s 1 s 2 a b a b a s 3 s 4 b V Goranko 6 of 21

Operations on binary relations: intersection R a ∩ b = R a ∩ R b s 1 s 2 s 3 s 4 V Goranko 7 of 21

Operations on binary relations: composition b a s 1 s 2 a a b b s 3 s 4 b a V Goranko 8 of 21

Operations on binary relations: composition R a ◦ b = R a ◦ R b b a s 1 s 2 a a b b s 3 s 4 b a V Goranko 9 of 21

Operations on binary relations: composition R a ◦ b = R a ◦ R b s 1 s 2 s 3 s 4 V Goranko 10 of 21

Operations on binary relations: composition R a ◦ b = R a ◦ R b s 1 s 2 s 3 s 4 V Goranko 11 of 21

Operations on binary relations: composition 2 b a s 1 s 2 a a b b s 3 s 4 b a V Goranko 12 of 21

Operations on binary relations: composition 2 R b ◦ a = R b ◦ R a b a s 1 s 2 a a b b s 3 s 4 b a V Goranko 13 of 21

Operations on binary relations: composition R b ◦ a = R b ◦ R a s 1 s 2 s 3 s 4 V Goranko 14 of 21

Operations on binary relations: composition 2 R b ◦ a = R b ◦ R a s 1 s 2 s 3 s 4 V Goranko 15 of 21

Auxiliary slides: reflexive and transitive closure of a relation Recall that, if W be a set and R ⊆ W 2 is a binary relation in W , then: R 2 = R ◦ R = { ( u , w ) | uRv and vRw for some v ∈ W } . Likewise, we define R 3 = R ◦ R 2 = R ◦ ( R ◦ R ), etc.... R n = R ◦ R n − 1 = R ◦ ( ... n times ) ... ◦ R ) ... ). Now, we define the reflexive and transitive closure of R as R ∗ := � n ≥ 0 R n , where R 0 = { ( w , w ) | w ∈ W } . Intuitively, R ∗ consists of all pairs ( u , v ) in W such that v can be reached from u in a finite number of R -steps. Formally, R ∗ = { ( u , v ) ∈ W 2 | uRu 1 ... Ru k = v for some u 1 , ..., u k ∈ W , k ≥ 0 } V Goranko 16 of 21

Operations on binary relations: transitive closure example R s 1 s 2 s 3 s 4 V Goranko 17 of 21

Operations on binary relations: transitive closure example R 2 s 1 s 2 s 3 s 4 V Goranko 18 of 21

Operations on binary relations: transitive closure example R ∪ R 2 s 1 s 2 s 3 s 4 V Goranko 19 of 21

Operations on binary relations: transitive closure example R 3 s 1 s 2 s 3 s 4 V Goranko 20 of 21

Operations on binary relations: transitive closure example R ∗ = R ∪ R 2 ∪ R 3 s 1 s 2 s 3 s 4 V Goranko 21 of 21