Strong LRSW assumption for type 3 pairing groups Why does the "strong LRSW" assumption by Ateniese et al. [Untraceable RFID Tags via Insubvertible Encryption, CCS'05] hold ONLY for type 3 pairings and NOT for symmetric pairings? Whereas, the LRSW assumption by Lysyanskaya et al. [Pseudonym Systems, SAC'99] does hold for all types of pairings.
In a type-3 group, after an oracle query you get to see $(g_1^a, g_1^{at}, g_1^{as+axst}, g_1^{ax}, g_1^{axt})$ but not $g_1^s$ or $g_1^t$ (those are in $G_2$). Suppose you had a homomorphism $G_2 \to G_1, g_2 \mapsto g_1$ so you could recover $g_1^s$ and $g_1^t$.

Call your tuple from the oracle $(A, B, C, D, E)$ and compute $(Ag_1^r, B(g_1^t)^r, C(g_1^s)^r, D, E)$ for an $r$ of your choice. There is a unique value of $u \in \mathbb Z_p$ (where $p$ is the group order) such that $a+r = ua$ and if my math is correct, the modified tuple is now a correct tuple for $a' = ua$ and $x' = u^{-1}x$, which the oracle never signed.

Indeed, for the $D, E$ components the change of variables $a \mapsto ua, x \mapsto u^{-1}x$ does not affect the product $a'x' = ax$ so we can leave them alone. For $C$, the exponent was $as + axst$, the second summand is ok but the first one has $as = a'u^{-1}s$ so we want to change that $a$ to $a'$. We can't do that directly, but since $au = a + r$ we can make up the difference by adding $g_1^{rs}$. This change from a multiplicative relationship (that leaves the product $ax$ alone) to an additive one (that we can use to adjust the remaining terms) is what makes this attack work. For $A, B$ we then do the appropriate adjustment as well.

Note that you never actually get to know the value of $u$, or $x'$. That's why this doesn't work against plain LRSW, where you need to output $x'$ as well.