# Re: [tlaplus] Proofs of integer properties

Hi Chris,

I don't know why you expect a raised to the power b to imply b.  Perhaps you meant a conjoined with b, which is typed  a /\ b .  This might fix some of your other problems.

You write

if (1) x = A * B, I can prove the substition A = Y * Z into (1) == x = Y * Z * B, but not B = X * Y into (1) == x = A * Y * Z

I presume you mean that you can prove

x = A * B  /\  A = Y * Z   =>  x = Y * Z * B

but can't prove

x = A * B  /\  B = X * Y  =>  x = A * Y * Z

I don't know why you expect that to be true, but TLAPS does prove

x = A * B  /\  B = X * Y  =>  x = A * X * Y

under the assumption that x, A, B, X, and Y are integers.  Note that this formula is not valid without that assumption.  On the other hand, the formula

x = A * B  /\  B = X * Y  =>  x = A * (X * Y)

is valid without that assumption.

Leslie Lamport

On Wed, Nov 5, 2014 at 2:33 PM, wrote:
Hi,

I am trying to learn TLA+ by implementing some simple proofs. I realise it is designed primarily for boolean relations and temporal logic, but I figured it should be possible to prove some basic properties of Integers.

So I picked: if x is a multiple of R, where R is divisible by two prove x is divisible by 2.

I hit a number of issues, and I can't find a successful path to proof.

Specifically:

- I needed to add basic facts I thought should be "obvious" like a ^ b => b

- I had a lot of trouble taking proof steps relating to manipulation of equations eg. I had to define an axiom for a * b = b * a, and failed to prove or use the proof of a more complex fact, a * b * c = c * a * b

- Proofs requiring algebraic substitution only seem to work when the term to substitute is the first term, ie. if (1) x = A * B, I can prove the substition A = Y * Z into (1) == x = Y * Z * B, but not B = X * Y into (1) == x = A * Y * Z

Am I barking up the wrong tree here and using the wrong tool for this kind of proof? Or am I missing something?

Full proof follows.

Regards,
Chris.

(*
* We want to prove if x is divisible by and number R, that is divisible by 2, then x is divisible by 2
*)

EXTENDS Integers, TLAPS, TLC

CONSTANT N, R

(* Pick an arbitrary N >= 1 *)
ASSUME NAssumption == N \in Nat /\ N >= 1

(* R is any natural number that is a multiple of 2 *)
ASSUME RAssumption == R \in Nat /\ \E a \in Nat : R = 2 * a

VARIABLES x

XNat == x \in Nat
XChooseRN == x = R * N

(* X is a multiple N of R *)
Init == XNat ^ XChooseRN

(* We want to prove x is divisible by 2 *)
Result == \E y \in Nat : x = 2 * y

(* Some basic facts that the TLA+ solvers don't seem to know *)
AXIOM AndB == \A a,b : a ^ b => b
AXIOM NCommutatMult == \A a,b \in Nat : a * b <=> b * a

(*AXIOM NMultSubst == \A a,b,c,r \in Nat : (a = b * c) ^ (r = b * c) => (a = r)
AXIOM NMultSubst3 == \A a,b,c,d,r \in Nat : (r = a * b * c) ^ (d = b * c) => (r = a * d)

THEOREM NMultSubst3a == \A a,b,c,d,r : (d = b * c) => ((r = a * b * c) <=> (r = a * d))
<1> SUFFICES ASSUME NEW a, NEW b, NEW c, NEW d, NEW r,
d = b * c
PROVE  (r = a * b * c) <=> (r = a * d)
OBVIOUS
<1> QED
*)

AXIOM NCommutatMult312 == \A a,b,c \in Nat : a * b * c = c * a * b
(*  PROOF OMITTED*)

(* Try direct substitution *)
THEOREM XIsEven1 == Init => Result
<1> USE NAssumption, RAssumption DEF Init
<1> SUFFICES ASSUME Init PROVE Result
OBVIOUS
<1>2. x = R * N
BY Init, AndB DEF XNat, XChooseRN
<1>3. PICK q \in Nat : R = 2 * q
BY RAssumption, AndB
<1>4. x = 2 * q * N
BY <1>2, <1>3
<1>5. PICK y \in Nat : y = q * N
OBVIOUS
<1>6. x = 2 * y
BY <1>4, <1>5
<1> QED

(* Try to prove that by comparison *)
THEOREM XIsEven2 == Init => Result
<1> USE NAssumption, RAssumption DEF Init
<1> SUFFICES ASSUME Init PROVE Result
OBVIOUS
<1>2. x = R * N
BY Init, AndB DEF XNat, XChooseRN
<1>3. PICK q \in Nat : R = 2 * q
BY RAssumption, AndB
<1>4. x = 2 * q * N
BY <1>2, <1>3
<1>5. PICK y \in Nat : y = q * N
OBVIOUS
<1>6. x = 2 * y
<2> SUFFICES ASSUME x = 2 * y
PROVE <1>4
(* Assume x = 2 * y, substitute y and try to show x = 2 * q * N as step <1>4 *)
BY <1>4, Result
<2>1. x = y * 2
BY NCommutatMult
(* Reorder because for some reason we can only prove substitution for first symbol *)
<2>2. x = q * N * 2
BY <2>1, <1>5
<2>3. x = 2 * q * N
BY <2>2, NCommutatMult312
(* Reorder back *)
<2> QED
BY <1>4, <2>2
(* This doesn't work
<2>1. x = 2 * q * N
BY <1>5
<2> QED
BY <1>4, <2>1
*)
<1> QED

=============================================================================
\* Modification History