# Does SARP imply WARP? and GARP imply SARP?

It’s obvious that WARP does not imply SARP, since WARP does not rule out cyclic choices, whereas SARP does.

The term “Strong” axiom suggests that it encompasses the “weak” axiom. But this is not obvious to me from the definition.

Does any choice structure that satisfies SARP also satisfy WARP?

Similarly, the "generalized" axiom of revealed preference (GARP) seems to encompass SARP, but again, I can't see this from the definition.

WARP: If $x,y\in B$, and $x\in C(B)$, then we cannot find a $B_2$ such that $x,y\in B_2$ and $y\in C(B_2)$ but not $x\in C(B_2)$.

SARP: Assume for all $B$ the choice $c(B)$ is only one element. If $x_i,x_{i+1}\in B_i$, and $x_i = c(B_i)$, for all $i\in \{1,N-1\}$, then $x_1=c(B_1)\notin B_N$.

GARP: if $p_i\cdot x_{i+1}\leq p_i\cdot x_i$ for $i\in \{1, N-1\}$ and $p_N\cdot x_{1}\leq p_N\cdot x_N$, then those inequalities must be equalities.

(GARP specifically assumes $p$ are price vectors and $x$ are consumption bundles, whereas the other two axioms apply to any type of choice structure.)

• Can you please add which versions of the axioms you use and what the domain is (budget sets, finite subsets of choices, etc)? – Michael Greinecker May 14 '18 at 7:14
• @MichaelGreinecker, Done, though please note that I am interested to know if there is a generalized version of GARP (one that doesn't just apply to competitive budget sets, but any type of choice structure, including non-market choice structures). – user56834 May 14 '18 at 8:18
• Also, I'm interested if there are alternative definitions that might change the conclusion. – user56834 May 14 '18 at 8:19

Yes, SARP implies WARP. Since all choices are singletons under SARP, we abuse notation and write $C$ as a function that takes alternatives as values. We prove the contrapositive, every violation of WARP in a setting where choices are unique give rise to a violation of SARP. So assume there are sets $B$, $B'$ containing $x$ and $y$ such that $c(B)=x$ and $C(B')=y$. Letting $B_1=B$ and $B_2=B'$, you get a direct violation of SARP (I take it you require $x_i\neq x_{i+1}$ in the definition of SARP).
In the context of budget sets, SARP implies GARP. Let $B_i=\{x\mid p_ix\leq p_ix_i\}$. Then SARP says that if $p_i x_{i+1}\leq p_i x_i$ but $x_i$ is the unique choice from $B_i$ for $i\in\{1,\ldots,N_1\}$, then $p_N x_i>p_N x_N$. So the case $p_N x_i\leq p_N x_N$ can never occur; GARP cannot be violated.