A simplified (κ,1)-morass is a structure M =<< θα | α ≤ κ >,< Fαβ | α < β ≤ κ >>
satisfying the following conditions:
(P0) (a) θ0 = 1, θκ = κ+, ∀α < κ 0 < θα < κ.
(b) Fαβ is a set of order-preserving functions f : θα → θβ.
(P1) |Fαβ| < κ for all α < β < κ.
(P2) If α < β < γ, then Fαγ = {f ∘ g | f Fβγ,g
Fαβ}.
(P3) If α < κ, then Fα,α+1 = {id ↾ θα,fα} where fα is such that fα ↾ δ = id ↾ δ and
fα(δ) ≥ θα for some δ < θα.
(P4) If α ≤ κ is a limit ordinal, β1,β2 < α and f1 Fβ1α, f2
Fβ2α, then there are a
β1,β2 < γ < α, g
Fγα and h1
Fβ1γ, h2
Fβ2γ such that f1 = g ∘ h1 and f2 = g ∘ h2.
(P5) For all α > 0, θα = ⋃
{f[θβ] | β < α,f Fβα}.
Lemma
Let α < β < κ, τ1,τ2 < θα, f1,f2 Fαβ and f1(τ1) = f2(τ2). Then τ1 = τ2 and
f1 ↾ τ1 = f2 ↾ τ2.
A simplified morass defines a tree < T,≺>.
Let T = {< α,ν >| α ≤ κ,ν < θα}.
For t =< α,ν > T set α(t) = α and ν(t) = ν.
Let < α,ν >≺< β,τ > iff α < β and f(ν) = τ for some f Fαβ.
If s ≺ t, then f ↾ (ν(s) + 1) is uniquely determined by the lemma. So we may define
πst := f ↾ (ν(s) + 1).
Lemma
The following hold:
(a) ≺ is a tree, htT (t) = α(t).
(b) If t0 ≺ t1 ≺ t2, then πt0t1 = πt1t2 ∘ πt0t1.
(c) Let s ≺ t and π = πst. If π(ν′) = τ′, s′ =< α(s),ν′ > and t′ =< α(t),τ′ >, then s′≺ t′ and
πs′t′ = π ↾ (ν′ + 1).
(d) Let γ ≤ κ, γ Lim. Let t
Tγ. Then ν(t) + 1 = ⋃
{rng(πst) | s ≺ t}.