Proofs on expressing nested GR(1) properties in GR(1)
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1 Proofs on expressing nested GR(1 properties in GR(1 Ioannis Filippidis Control and Dynamical Systems California Institute of Technology December 22, 2017 Abstract This is a proof that any nested GR(1 property is equivalent to a GR(1 property, by adding an auxiliary variable that counts up to the nesting depth, which is at most equal to the number of states. In practice the nesting depth is small. The proof style below is described in [1]. The results are proved for stutter-sensitive properties, and stutter-sensitive temporal quantification (,. Wherever we invoke arguments about history-determined variables, these are analogous to the stutter-invariant case [2]. The rules of TLA + are assumed for declaring variables, so when writing q : a in a theorem statement, the identifier q stands for a fresh name in the context where the theorem is applied. For this reason, where possible we do not mention assumptions about what identifiers can occur when. assume The operators take Boolean values, except where otherwise noted. Definition 1 Let d Nat, H k Nat, Ξ m Nat. A chain condition is a formula of the form: m 1..d : P m 1 Q m l 0..Ξ m : ξ ml P m 1 m 0..d : Q m P m l 0..H m : η ml (P m Q m l 0..Ξ m : ξ ml Q m assume ChainCondition = The operators P m, Q m, ξ ml, η ml satisfy a chain condition (Definition 1. Definition 2 (Nested GR(1 property [3] Let φ is a nested GR(1 property. φ = m 0..d P m k m..d Q m ( Q m l 0..Ξ m 1
2 A nested GR(1 property is defined here as a liveness formula. Inserting a liveness formula of this form in a stepwise implication operator ( +, WhilePlusHalf, or some other choice can yield an open-system property. Proposition 3 let φ 1 = (( i A i B = ( i A i B in φ define φ 1 = (( i A i B = ( i A i B 1 2. suffices assume new σ, IsABehavior(σ prove σ = φ case σ = i A i 2 1. σ = ( i A i ( i A i 2 2. case σ = B 3 1. σ = B BY σ = i A i by 1 3, σ = φ 1 ( i A i B ( B ( i A i ( i A i B 3 4. σ = B i A i ( ( i A i B 3 5. qed by 3 3, case σ = B by 2 3 def φ 1, 2 4. qed by 2 2, case σ = i A i 2 1. σ = φ 1 by 1 4 def φ σ = 3 1. σ = i A i by qed by 3 1 def 2 3. qed by 2 1, qed by 1 3, 1 4 2
3 Proposition 4 prove let φ 1 in φ 1 = (a b = q : q = true ( q (b (q a q 1 1. suffices assume new σ, IsABehavior(σ prove σ = φ assume σ = φ 1 prove σ = The variable q is history-determined in Hist q : (q = true (q = b ( a q. It remains to show that φ 1 and Hist imply q. The case a implies b, so q. The case a is split as ( a (a a. For the subcase a, q starts and remains true. For the subcase (a a, some infinite suffix of σ has a from the second state onwards, and b, so q. Thus, q assume σ = prove σ = φ 1 The case (a b implies a b. The case (a b implies q. By q, eventually there is a (q q-step. Such a step implies b. The two cases are exhaustive, thus (a b qed by 1 2, 1 3 Proposition 5 prove ( ((a j b j c ( ( j b j (a c 1 1. ( (a j b j c ( ( j b j (a c 1 2. qed by 1 1, Proposition 3 with A i b j, i j, B (a c Proposition 6 prove ( ( j b j (a c q : ( j b j q = true (q = c (q a q proof by Proposition 4, rules for. From this point on we assume that history-determined variables (q, r, etc. do not occur implicitly (so c, a, etc. below are independent of those history variables, only explicitly where shown. 3
4 Proposition 7 define φ 1 = q : ( j b j q = true (q = c ( a q q = r, q : ( j b j q = true r true (r = c (r a (q = c (q a q prove φ 1 proof by the rules for introducing bound variables, r does not occur in any of the expressions a, c, b j, q. So r is a history-determined variable. Proposition 8 define φ 1 = r, q : ( j b j q = true (q = c ( a q q r = true (r = c ( a r = r : ( j b j r r = true (r = c ( a r prove φ φ 1 Within the consequent, q and r are both history-determined by the same formulas. Thus, (q = r, which together with q implies r in the consequent. We then eliminate q (recall the assumption that q, r do not occur in a, c, b j φ 1 Introduce a history-determined variable r in the consequent. Again, identical formulas imply that (r = q within the consequent, so q implies r. Omit the conjunct q qed by 1 1, 1 2 Proposition 9 prove ( ( (a j b j c q : q = true (q = c ( a q ( j b j q (1 by Proposition 5, Proposition 6, Proposition 7, Proposition 8. 4
5 Proposition 10 assume For all m 0..d, for any j with j 0..d j m, variable q m does not occur in any of P m, η kl, Q m. prove let φ 1 = m 0..d ((P m ( k m..d Q m = q 0,..., q d : m 0..d q m true in φ 1 ( q m = Q m ( P m q m ( k m..d q m 1 1. assume new m 0..d prove ((P m ( k m..d Q m q m : q m (q m = Q m ( P m q m ( k m..d q m proof Substitute a P m, b j η kl, c Q m, q q m in Proposition qed By hypothesis, 1 1, dm=0 ((P m ( Q m k m..d dm=0 q m : q m true q 0,..., q d (q m = Q m ( P m q m ( k m..d q m : d m=0 q m true (q m = Q m ( P m q m ( k m..d q m (2 Proposition 11 prove let φ 1 in φ 1 = m 0..d q m = true 1 1. define φ 1 = m 0..d q m = true ( q m = Q m ( P m q m ( k m..d = m 0..d q m = true ( q m = Q m ( P m q m ( k 0..d ( q m = Q m ( P m q m ( k m..d = m 0..d q m = true ( q m = Q m ( P m q m ( k 0..d 1 2. suffices assume new σ, IsABehavior(σ prove σ = φ 1 q m q m q m q m 5
6 1 3. assume σ = φ 1 prove σ = obvious 1 4. assume σ = prove σ = φ k 0..d : l 0..H k : η kl (P k Q k P k 2 2. k 0..(d 1 : P k Q k k 0..(d 1 : l 0..H k : η k P k Q k+1 by 2 1, k 0..(d 1 : r k..d : Q k Q r 3 1. k 0..(d 1 : Q k Q k k 0..(d 1 : P k Q k+1 by ChainCondition 4 2. k 0..d : Q k P k 4 3. qed by 4 1, k 0..(d 1 : r (k + 1..d : Q k Q r 4 1. r (k + 1..d : k < k + 1 r d k < r d 4 2. qed by 4 1, 3 1, NatInduction 3 3. k : Q k Q k 3 4. qed by 3 2, m 0..d : k 0..(m 1 : l 0..H k : η kl Q m 3 1. m 0..d : k 0..(m 1 : l 0..H k : η kl Q k m 0..d : k 0..(m 1 : k 0..(m 1 = 0..(d qed by 4 1, m 0..d : k 0..(m 1 : Q k+1 Q m 4 1. m 0..d : k 0..(m 1 : m (k + 1..d 5 1. k 0..(m 1 k m 1 k + 1 m 5 2. m 0..d m d 5 3. qed by 5 1, m 0..d : k 0..(m 1 : (k (m 1+1 = 0..m 0..d (the k +1 here is k in qed by 4 1, 4 2, qed by 3 1, 3 2, m 0..d : k 0..(m 1 : l 0..H k : ( (η kl Q k+1 (Q k+1 Q m η kl Q m 2 6. m 0..d : k 0..(m 1 : l 0..H k : η kl Q m 3 1. (A B ( A B 3 2. ( B ( B 3 3. (A B ( A B by 3 1, (η kl Q m ( η kl Q m by 3 3 with A η kl, B Q m 6
7 3 5. qed by 2 5, σ = m 0..d ( q m = Q m ( P m q m 2 8. m 0..d : σ = Q m q m 3 1. σ = Q m q m by qed by 3 1 and q m q m 2 9. m 0..d : k 0..(m 1 : l 0..H k : σ = η kl q m by 2 6, m 0..d : σ = ( k 0..(m m 0..d : σ = ( k 0..(m 1 by m 0..d : ( k 0..(m qed by 3 1, qed by 2 10, as follows: 1 5. qed by 1 3, 1 4 Proposition 12 let φ 1 η kl q m η kl k 0..(m 1 q m q m σ = σ = m 0..dq = true m 0..d (q m = Q m ( P m q m k 0..d η kl q m σ = m 0..dq = true m 0..d (q m = Q m ( P m q m k m..d η kl k 0..(m 1 η kl q m σ = m 0..dq = true m 0..d (q m = Q m ( P m q m k m..d η kl q m q m σ = m 0..dq = true m 0..d (q m = Q m ( P m q m k m..d η kl q m σ = φ 1 = m 0..d P m k m..d Q m q m = q 0,..., q d : m 0..dq m true m 0..d ( q m Q m ( P m q m ( η ml m 0..d q m in φ 1 7
8 by Propositions 10 and 11 Proposition 13 assume new m 0..d prove let φ 1 = ( Q m l 0..Ξ m = l 0..Ξ m in φ define φ 1 = ( Q m l 0..Ξ m = l 0..Ξ m 1 2. suffices assume new σ, IsABehavior(σ prove σ = φ assume σ = prove σ = φ Ξ m l=0 Ξ m l=0 proof Ξ m l=0 Ξ m l=0 Ξ m l= ( Ξ m l=0 (( Q m Ξ m l=0 proof ( Ξ m l=0 (( Q m Ξ m l= qed by 1 3, 2 1, assume σ = φ 1 prove σ = 2 1. case σ = Q m 3 1. l 0..Ξ m : ξ ml Q m by ChainCondition 3 2. l 0..Ξ m : Q m ξ ml by l 0..Ξ m : ( Q m ( by qed by 2 1, case σ = Q m 3 1. φ 1 Q m l 0..Ξ m 3 2. l 0..Ξ m l 0..Ξ m l 0..Ξ m by [4, p.93], (A B ( A ( B qed by 2 2, 3 1, qed by 1 4, 2 1, qed by 1 3, 1 4 Proposition 14 prove let φ 1 = ( η ml m 0..d = m 0..d l 0..Ξ m in φ 1 l 0..Ξ m 8
9 1 1. define φ 1 = ( η ml m 0..d l 0..Ξ m = m 0..d l 0..Ξ m 1 2. suffices assume new σ, IsABehavior(σ prove σ = φ σ = φ 1 obvious 1 4. σ = φ define F = η ml 2 2. case σ = F by 2 2 def F, φ 1, 2 3. case σ = F 3 1. ( F η ml 3 2. η ml m 0..d : l 0..H m : η ml 3 3. m 0..d : l 0..H m : η ml k 0..d : r 0..H k : ξ kr 4 1. m 0..d : l 0..H m : η ml P m Q m 4 2. m 0..(d 1 : P m P m m 0..(d 1 : q (m + 1..d : P m P q by 4 2, NatInduction 4 4. m 0..(d 1 : q (m + 1..d : P q P m 4 5. m 0..d : l 0..Ξ m : ξ ml Q m by ChainCondition 4 6. m 1..d : l 0..Ξ m : ξ ml P m 1 by ChainCondition 4 7. ξ ml P m 1 P m 2 P 0 r 0..(m 1 : l 0..Ξ m : ξ ml P r and m (r + 1..d : l 0..Ξ m : P r ξ ml 4 8. m 0..(d 1 : Q m Q m m 0..(d 1 : q m..d : Q q Q m and Q q Q q m 0..d : l 0..Ξ ml : ξ ml Q m proof m 0..d : l 0..Ξ m : Q m ξ ml m 0..q : l 0..Ξ m : Q q ( Q m ξ ml qed proof m 0..d : l 0..H m : η ml Q m k 0..m : θ 0..Ξ k : ξ kθ by 4 1, Also, m 0..d : l 0..H m : η ml P m k (m + 1..d : θ 0..Ξ k : ξ kθ by 4 7. These imply that k 0..d : θ 0..Ξ k : ξ kθ m 0..d : l 0..H m : η ml k 0..d : r 0..Ξ k : ξ kr by m 0..d : l 0..H m : η ml k 0..d : r 0..Ξ k : ξ kr by 3 4, ( A ( A 3 6. m 0..d : l 0..H m : η ml k 0..d 3 7. η ml k 0..d by 3 2, qed by 2 3, 3 7 def φ 1, 2 4. qed by 2 2, qed r 0..Ξ k ξ kr r 0..Ξ k ξ kr 9
10 P m Q m q m q m Q m Q m Q m P m Figure 1: The automaton that corresponds to the auxiliary variable p that reduces memory to d + 1. by 1 3, 1 4 Proposition 15 prove ( m 0..d (Q m l 0..Ξ m η ml m 0..d l 0..Ξ m by Propositions 13 and 14 Lemma 16 let φ 1 in φ 1 by Propositions 12 and 15 = m 0..d P m k m..d Q m ( Q m l 0..Ξ m = q 0,.., q d : m 0..d q m true ( q m Q m ( P m q m η ml m 0..d q m l 0..Ξ m 10
11 Proposition 17 prove let φ 1 in φ 1 = q 0,.., q d : m 0..d q m = true ( q m = Q m ( P m q m q m = p : p = d + 1 p = choose r 0..(d + 1 : m 0..d : (m < r Q m P m p > m (r d Q r P r p > r (p = d + 1 below is stutter-sensitive temporal quantification. Respose(q0,.., qd = m 0.. d : qm true (qm = Qm ( Pm qm qm phi1 = q0,.., qd : Response(q0,.., qd ChooseP(r = m 0.. d : (m < r Qm (r d Qr Pr (p > r Pm (p > m SMP(p = p = d + 1 (p = choose r 0.. (d + 1 : ChooseP(r SpecP(p = SMP(p (p = d + 1 phi2 = p : SpecP(p The satisfaction relation ( = below is of raw TLA+ theorem ReductionWithLinearMemory = assume new sigma, IsABehavior(sigma, ChainCondition prove sigma = phi1 phi2 proof 1 1. sigma = phi1 phi suffices assume sigma = phi1 11
12 prove sigma = phi2 obvious 2 2. sigma = p : SMP(p by def SMP p is history-determined we will prove liveness of p in SpecP 2 3. pick tau : IsABehavior(tau EqualUpToVars(tau, sigma, p, q0,.., qd tau = SMP(p Response(q0,.., qd by suffices tau = (p = d + 1 by 2 3 def SpecP, SMP, phi2 and that p does not occur in the spec 2 9. tau = ( r 0.. (d + 1 : ChooseP(r 3 1.case m 0.. d : Qm ( Pm (p > m 4 1. m 0.. d : (m < (d + 1 (Qm ( Pm (p > m by ChooseP(d + 1 (d (d + 1 by 4 1 def ChooseP 4 qed by case m 0.. d : Qm (Pm (p > m 4 1. pick z 0.. d : Qz (Pz (p > z t 0.. d : (t < z (Qt ( Pt (p > t by 3 2, SmallestNumberPrinciple choose the smallest such z (z d ( Qz (Pz (p > z by ChooseP(z z 0.. (d + 1 by 4 1, qed by r 0.. (d + 1 : ChooseP(r by 3 1, 3 2 which are exhaustive 3 qed by tau = p 0.. (d + 1 ChooseP(p by 2 3, 2 9 def SMP The next step ensures that if ever p < (d + 1, then qm, which starts a chain of Qm, leading to the bottom, and to (p = d + 1 again tau = p p m 0.. d : p = m qm 12
13 by 2 10, 2 3 (p 0..(d + 1, so if p decreases, then p = r 0..d. Since Choose(p, it is (Qr ( Pr (p > r, thus qr, by def Response (qm boolean (qm = Qm ( Pm qm (( qm (Qm qm by PTL 2 7. tau = (( qm qm Qm (( qm (Qm qm 3 1. tau = qm = true (qm = Qm ( Pm qm by 2 1, 2 3 def EqualUpToVars, phi 1 3 qed by 3 1, tau = m 0.. d : (( qm Qm 3 1. tau = m 0.. d : qm by 2 3 def Response 3 qed by 3 1, tau = (p = m (p 0.. d Qm p = m 3 1. Qm P(m 1.. P0 by ChainCondition 3 2. (p = m p 0.. d r 0.. (m 1 : (p > r obvious 3 3. ( Qm (p = m (p 0.. d r 0.. (m 1 : p > r Pr by 3 1, tau = ( ( Qm (p = m (p 0.. d (p m by 3 3, 2 10 def ChooseP Otherwise the second conjunct of ChooseP would be violated ( Qm (p = m (p 0.. d ( Qm (p m obvious 3 6. tau = ( ( Qm (p = m (p 0.. d (p m by 3 5 def ChooseP Otherwise (m < p so it must be (Qm ( Pm (p > m, which is not the case. 3 qed by 3 4, define TopOrLower(m = (0.. (d + 1 \ (m.. d 13
14 2 12. tau = (Qm qm p TopOrLower(m 3 1. tau = (Qm qm by 2 3 def Response 3 2. Qm Q(m Qd by ChainCondition 3 3. tau = (Qm (p / m.. d by 2 10, 3 2 def ChooseP The second conjunct of ChooseP would be violated by such a p as r tau = (p 0.. (d + 1 by 2 10, RawRuleINV tau = (Qm (p ((0.. (m 1 {d + 1} by 3 3, qed by 3 1, 3 5 def TopOrLower 2 qed by 2 5, 2 7, 2 11, 2 12 Initially p = d + 1 by def SMP. Assume p < p. Then by 2 5, m 0..d : (p = m qm. It is qm Qm. Until qm, it is ( qm qm. by 2 7 follows that Qm until qm qm. by 2 7, qm qm implies Qm. by 2 11, p = m = p while Qm, so while qm qm. by 2 12 Qm implies (p < p = m (p = d + 1. So ((p < p Qm (because we started with Qm, and Qm. So p is strictly decreasing until p = d + 1 (by induction over steps of the behavior sigma = phi2 phi suffices assume sigma = phi2 prove sigma = phi1 obvious 2 2. sigma = p : SpecP(p by 2 1 def phi sigma = q0,.., qd : m 0.. d : qm = true (qm = Qm ( Pm qm obvious q 0,..., q d are history-determined pick tau : IsABehavior(tau EqualUpToVars(tau, sigma, p, q0,.., qd tau = SpecP(p m 0.. d : qm = true (qm = Qm ( Pm qm by 2 2, suffices tau = m 0.. d : qm by 2 4 def phi 1, Response 2 6. suffices tau = m 0.. d : ((qm qm qm by 2 4, PTL whenever any q m becomes false, it eventually becomes again true. Thus, the goal from 2 5 follows tau = m 0.. d : ((qm qm ( Qm Pm by tau = m 0.. d : (p > m 14
15 by 2 4 def SpecP 2 9. tau = m 0.. d : (( Qm Pm (p m by 2 4 def SpecP, SMP, ChooseP we proved under 1 1 that ChooseP(p holds tau = m 0.. d : ((p > m (Qm ( Pm (p > m by 2 4 def SpecP, SMP, ChooseP tau = m 0.. d : (( Qm Pm ((p m (p > m by 2 9, tau = m 0.. d : (( Qm Pm Qm by 2 10, tau = m 0.. d : (Qm qm by tau = m 0.. d : (( Qm Pm qm by 2 12, qed by 2 7, qed by 1 1, 1 2 References [1] L. Lamport, How to write a 21st century proof, Journal of fixed point theory and applications, vol. 11, no. 1, pp , [2] M. Abadi and L. Lamport, An old-fashioned recipe for real time, TOPLAS, vol. 16, no. 5, pp , September [3] I. Filippidis and R. M. Murray, Symbolic construction of GR(1 contracts for systems with full information, in ACC, 2016, pp [4] L. Lamport, Specifying Systems: The TLA + language and tools for hardware and software engineers. Addison-Wesley,
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