Metamath Proof Explorer


Theorem dfsuccf2

Description: Alternate definition of Scott Fenton's version of Succ , cf. df-sucmap . (Contributed by Peter Mazsa, 6-Jan-2026)

Ref Expression
Assertion dfsuccf2 Succ = { ⟨ 𝑚 , 𝑛 ⟩ ∣ suc 𝑚 = 𝑛 }

Proof

Step Hyp Ref Expression
1 df-succf Succ = ( Cup ∘ ( I ⊗ Singleton ) )
2 df-co ( Cup ∘ ( I ⊗ Singleton ) ) = { ⟨ 𝑚 , 𝑛 ⟩ ∣ ∃ 𝑥 ( 𝑚 ( I ⊗ Singleton ) 𝑥𝑥 Cup 𝑛 ) }
3 vex 𝑚 ∈ V
4 vex 𝑛 ∈ V
5 3 4 lemsuccf ( ∃ 𝑥 ( 𝑚 ( I ⊗ Singleton ) 𝑥𝑥 Cup 𝑛 ) ↔ 𝑛 = suc 𝑚 )
6 eqcom ( 𝑛 = suc 𝑚 ↔ suc 𝑚 = 𝑛 )
7 5 6 bitri ( ∃ 𝑥 ( 𝑚 ( I ⊗ Singleton ) 𝑥𝑥 Cup 𝑛 ) ↔ suc 𝑚 = 𝑛 )
8 7 opabbii { ⟨ 𝑚 , 𝑛 ⟩ ∣ ∃ 𝑥 ( 𝑚 ( I ⊗ Singleton ) 𝑥𝑥 Cup 𝑛 ) } = { ⟨ 𝑚 , 𝑛 ⟩ ∣ suc 𝑚 = 𝑛 }
9 1 2 8 3eqtri Succ = { ⟨ 𝑚 , 𝑛 ⟩ ∣ suc 𝑚 = 𝑛 }