𝑡𝑟𝑢𝑒⇔{𝑡𝑟𝑢𝑒⇔∅⊆{𝗍𝗋𝗎𝖾,𝖿𝖺𝗅𝗌𝖾,0}⇔S0⊆S1∀i∈ℕ(Si⊆Si+1⇒∀t∈Si+1{t∈{𝗍𝗋𝗎𝖾,𝖿𝖺𝗅𝗌𝖾,0}∨∃t1∈Sit∈{𝗌𝗎𝖼𝖼 t1,𝗉𝗋𝖾𝖽 t1,𝗂𝗌𝗓𝖾𝗋𝗈 t1}∨∃t1∈Si,t2∈Si,t3∈Sit=𝗂𝖿 t1𝗍𝗁𝖾𝗇 t2𝖾𝗅𝗌𝖾 t3∀t1∈Si(𝑡𝑟𝑢𝑒⇔t1∈Si+1⇒{𝗌𝗎𝖼𝖼 t1,𝗉𝗋𝖾𝖽 t1,𝗂𝗌𝗓𝖾𝗋𝗈 t1}⊆Si+2)∀t1∈Si,t2∈Si,t3∈Si(𝑡𝑟𝑢𝑒⇔{t1,t2,t3}⊆Si+1⇒(𝗂𝖿 t1𝗍𝗁𝖾𝗇 t2𝖾𝗅𝗌𝖾 t3)∈Si+2)⇒Si+1⊆Si+2)⇒∀i∈ℕSi⊆Si+1
sketch of proofs:
∀s∈S(∀r∈S(𝑑𝑒𝑝𝑡ℎ r<𝑑𝑒𝑝𝑡ℎ s⇒Pr)⇒Ps)⇔∀n∈ℕ∗(∀r∈S(𝑑𝑒𝑝𝑡ℎ r<n⇒Pr)⇒∀s∈S(𝑑𝑒𝑝𝑡ℎ s=n⇒Ps))⇔∀n∈ℕ∗(∀m∈ℕ∗,m<n,s∈S(𝑑𝑒𝑝𝑡ℎ s=m⇒Ps)⇒∀s∈S(𝑑𝑒𝑝𝑡ℎ s=n⇒Ps))⇒∀n∈ℕ∗,s∈S(𝑑𝑒𝑝𝑡ℎ s=n⇒Ps)⇔∀s∈SPs
∀s∈S(∀r∈S(𝑠𝑖𝑧𝑒 r<𝑠𝑖𝑧𝑒 s⇒Pr)⇒Ps)⇔∀n∈ℕ∗(∀r∈S(𝑠𝑖𝑧𝑒 r<n⇒Pr)⇒∀s∈S(𝑠𝑖𝑧𝑒 s=n⇒Ps))⇔∀n∈ℕ∗(∀m∈ℕ∗,m<n,s∈S(𝑠𝑖𝑧𝑒 s=m⇒Ps)⇒∀s∈S(𝑠𝑖𝑧𝑒 s=n⇒Ps))⇒∀n∈ℕ∗,s∈S(𝑠𝑖𝑧𝑒 s=n⇒Ps)⇔∀s∈SPs
∀s∈S(∀r(r is a subterm of s⇒Pr)⇒Ps)⇒∀i∈ℕ(∀j∈ℕ,j<i,s∈SjPs⇒∀s∈SiPs)⇒∀i∈ℕ,s∈SiPs⇒∀s∈SPs