plexity
• For the direction from left to right.
Assume φis satisfiable in PL. Let V be the valuation evaluationφ to1. Define a MPl valuationV′ as follows:
V′(p)(k) =w iffV(p) =1for all playersk6= 1,
V′(p)(1) =w iffV(p) =0.
By induction onφwe will show that player 0 has a winning strategy for the gameG(φ′, V′).
– Base case.
- φ=p. In this caseφ′ =∨
1{∼0k p|k∈Ag− {1}}. SinceV(p)
must be1V′(p)(k) =wfor allk6= 1. Thus, whichever conjunct
player 1 will choose, 0 will win. Hence, 0 has a winning strategy for the gameG(φ′, V′).
- φ=⊤, thenφ′ =⊤
0 and 0 has a winning strategy for the game
G(φ′, V′).
- φ = ¬p, then φ′ =∼
ij p. Since V(p) = 0 it follows that
V′(p)(1) =w, and 0 has a winning strategy.
- φ=¬⊥, thenφ′ =∼01 ⊥0. It follows immediately that 0 has a
winning strategy for the gameG(φ′, V′). – Induction step.
- φ= ψ1∨ψ2, then φ′ is of the form ψ1′ ∨iψ′2. Since V(φ) = 1
it follows that there is an l ∈ {1,2} such that V(ψl) = 1. By
induction hypothesis it follows that 0 has a winning strategy for G(ψ′
l, V′) and thus she also has a winning strategy forG(φ′, V′).
(She can simply chooseψ′
lin her first move, then play according
to her winning strategy).
- φ = ψ1∧ψ2, then φ′ is of the form ψ′1∨1ψ2′.Since V(φ) = 1
it follows that for both l ∈ {1,2}, V(ψl) = 1. By induction
hypothesis it follows that 0 has a winning strategy forG(ψ′
l, V′)
for bothl= 0 and l = 1. Thus she also has a winning strategy for G(φ′, V′). (Whichever conjunct 1 chooses she can use her
winning strategy for that conjunct).
• For the direction from right to left.
Assume φ′ is 0-satisfiable using restricted valuations. LetV′ be the val-
uation such that player 0 has a winning strategy for the game G(φ′, V′).
Define aPLvaluationV as follows:
V(p) =1iff there is ak6= 1 such thatV′(p)(k) =w,
By induction onφwe will show thatV′(φ′)(0) =w. – Base case.
- φ = p. In this case φ′ =∨
1{∼0k p| k ∈ Ag− {1}}. Since 0
has a winning strategy forG(φ′, V′) it follows that for all agents
k6= 0, V′(p)(k) =w. Hence, V(p) =1.
- φ=⊤, thenφ′ =⊤0 andV(⊤) =1.
- φ=⊥, then φ′=⊥
0, which is not 0-satisfiable, henceφcannot
equal⊥.
- φ=¬p, thenφ′ =∼
ij p. Since 0 has a winning strategy it follows
thatV′(p)(1) =w, henceV(p) =1andV(¬φ) =1.
- φ=¬⊤, then φ′ =∼
01 ⊤0 which equals⊤1. Since the formula
⊤1 is not 0-satisfiableφcannot equal ¬⊤.
- φ=¬⊥, thenφ′ =∼
01⊥0. It follows immediately thatV(¬⊥) = 1.
– Induction step.
- φ = ψ1∨ψ2, then φ′ is of the form ψ1′ ∨i ψ′2. Since 0 has a
winning strategy forG(φ′, V′) it follows that there is anl∈ {1,2}
such that 0 has a winning strategy for G(ψ′
l, V′). By induction
hypothesis it followsV(ψl) =1, whenceV(φ) =1.
- φ=ψ1∧ψ2, thenφ′is of the formψ1′∨1ψ2′.Since 0 has a winning
strategy forφ′ it follows that for bothl∈ {1,2}, 0 has a winning
strategy. By induction hypothesis it follows that both V(psi1)
andV(ψ2) evaluates to 1, henceV(φ) =1.
Definition 7.4.1 (Translation of the formula φ(m) forcing the existence of binary trees). For any proposition letterpdefinep=W1{∼0kp|k∈A− {1}}.
The formulaφ′(m)will be the ∨
1-conjunction of the following formulas:
(i) q0 (ii) ♦m 1(∼01qi∨0(W1{∼01qj | forj6=i}))for0≤i≤m (iii) B0∨1♦1B1∨1♦21B2∨1. . .∨1♦m1−1Bm−1 (iv) ♦1S(p1,∼01p1)∨1♦21S(p1,∼01p1). . .∨1♦1m−1S(p1,∼01p1) ∨1♦21S(p2,∼01p2). . .∨1♦m1−1S(p2,∼01p2) .. . ♦m1−1S(pm−1,∼01pm−1),
Bi=∼01qi∨0(♦0(qi+1∨1pi+1)∨1♦i(qi+1∨1∼01pi+1)),
and,
S(pi,∼01pi) = (∼01pi∨0♦1pi)∨1(pi∨0♦1∼01pi).
This formula φ′(m) is 0-satisfiable. However, it forces its models to be ex-
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