Following up on Saksham's reply: integer division is written \div in TLA+ and is supported by the prover. Proving theorems about an _expression_ CHOOSE x \in S : P(x) is more complicated because it usually requires showing that there is some element of S that satisfies P. Since you mention min and max, below is a proof by induction over finite sets that shows that any finite and non-empty set of integers has a maximum element. (Your module needs to extend the modules FiniteSetTheorems, which contains FS_Induction and other useful theorems about finite sets and TLAPS.)
Assuming that you are more interested in proving theorems about an algorithm than about elementary lemmas about data, you may want to leave such lemmas unproved.
Stephan
Max(S) == CHOOSE x \in S : \A y \in S : y <= x
LEMMA MaxIntegers == ASSUME NEW S \in SUBSET Int, S # {}, IsFiniteSet(S) PROVE /\ Max(S) \in S /\ \A y \in S : y <= Max(S) <1>. DEFINE P(T) == T \in SUBSET Int /\ T # {} => \E x \in T : \A y \in T : y <= x <1>1. P({}) OBVIOUS <1>2. ASSUME NEW T, NEW x, P(T), x \notin T PROVE P(T \cup {x}) <2>. HAVE T \cup {x} \in SUBSET Int <2>1. CASE \A y \in T : y <= x BY <2>1, Isa <2>2. CASE \E y \in T : ~(y <= x) <3>. T # {} BY <2>2 <3>1. PICK y \in T : \A z \in T : z <= y BY <1>2 <3>2. x <= y BY <2>2, <3>1 <3>3. QED BY <3>1, <3>2 <2>. QED BY <2>1, <2>2 <1>. HIDE DEF P <1>3. P(S) BY <1>1, <1>2, FS_Induction, IsaM("blast") <1>. QED BY <1>3, Zenon DEF Max, P
I'm able to use the above operator with Z3. Thank you for your help.
Hans The following should work and invoke Z3:
divide_by_two(n) == n \div 2
Hi,
I'm working on proving an algorithm using TLAPS that requires dividing integers by two. I understand there's no support for real numbers/division in the backend provers, so I am using an operator:
divide_by_two(n) == CHOOSE k \in Nat: n=2*k
The problem I'm running into is when trying to prove properties using the above operator with set expressions. The following example demonstrates this:
EXTENDS Integers, TLAPS, FiniteSets
divide_by_two(n) == CHOOSE k \in Nat: n=2*k
CONSTANTS N
VARIABLES x
Init == x = divide_by_two(N)
PositiveInvariant == x >= 0
ASSUME NumberAssumption == N \in {2,4,6,8,10}
THEOREM PositiveDivisionProperty == Init => PositiveInvariant <1> SUFFICES ASSUME Init PROVE PositiveInvariant OBVIOUS <1> QED BY NumberAssumption DEF Init, PositiveInvariant, divide_by_two
The backend provers are unable to prove the above theorem. SMT specifically times out and increasing the timeout doesn't appear to help.
Changing the assumption to:
ASSUME NumberAssumption == N = 10
will successfully prove the theorem. Are the backend provers able to work with 'CHOOSE' _expression_ in conjunction with sets? Is there anything wrong with how I'm specifying the assumption? The actual algorithm will require using the division operator with a few 'min'/'max' operators, which may also complicate the proof process, so I'm wondering if TLAPS is able to prove these sort of algorithms that require arithmetic.
Best, Hans
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