The name is absent

v_A(k + 2) + v_b(к + 2) — Z'Z_a = Z 1 — δ)vB(к + 1) > 0. However, by (9)
v_A(k + 2) + v_b(к + 2) ≤ max(δ^k+2Z_a.Z" ^(k'²Z_b). Moreover, by definition of
∕₀,Z" ⁱ Z_a > Z" ^u" ⁱ'Z_b, which, since к + 2 ≤ J₀, implies that δ^k'^{2 i}Z_a >
Z" ^(k'²''1 Z_a, or δ^kZa > δ^m~(^k+2)ZB. This in turn implies that v_A(k + 2) +
v_b (k + 2) ≤ max(δ^k+2Z_a.Z" ^(k'² Z_b) < δ^kZ_a.contradicting the claim that
v_A(k + 2) + v_b(k + 2) — δ^kZ_a = Z ¹ — δ')v_B(k + 1) > 0 Hence, v_b(k + 1) = 0.
which immediately implies that v_A(k) = δv_A(k — 1) = δ^kZ_a.

This induction argument therefore shows for all k ≤ J₀ — 1, vb(k) = 0
and for all k ≤ J₀ — 2.v_A(k) = δ^kZ_a . An immediate consequence is that

^VA^(J0 ^{— 1}) = ^δ[(^vA(^j0 ^{— 2) + v}B ^(j0 ^{— 2)) — v}B ^(j0^)] = ^δf^δ7" 2^ZA ^{— v}B ^(j0^{)] (25)}

This equation will be used in the continuation to derive values at J₀ — 1 and
^j0.

To address equilibrium behavior in states greater than or equal to J₀.we
start in state m. Note that since m is a terminal state v_b (m) = Z_b and
v_A(m) = 0.If J₀ = m — 1. then by definition δ^' ⁱZ_a < δZ_B and δ^' ²Z_a >

δ²Z_B. Moreover, by Claim 1 v_b(m—2) = 0 and by (25) v_A(m—2) = δ^' ²Z_a —
δv_B (m — 1). Hence,

v_A(m — 1) = δv_A(m) + δ max(0.v_A(m — 2)+ v_b (m — 2) — v_A(m) — v_b (m))

= 0 + δ max(0.J (m — 2) — Z_b )

≤ δmax(0. niaxZ”' ²Z_aZ²Z_a) — Z_b)

≤ 0

where the final inequality follows from the fact that δ"' ²Z_a = δ^j" ⁱZ_a <
δm-j"-¹ZB = δ^m-(^m-1-¹ZB = Zb. Hence, VA(m — 1) = 0. It follows that

v_b (m—1) = δv_B (m—2)+δ max(0.Z_B —v_A(m—2)—v_b (m—2)) = δ(Z_B—v_A(m—2))

Hence,

v_A(m — 2) = Z" ²Z_a — δv_B (m — 1) and

vb (m — 1) = δ(ZB — VA(m — 2))

Solving simultaneously implies

25

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