Metamath Proof Explorer


Theorem dfsuccl4

Description: Alternate definition that incorporates the most desirable properties of the successor class. (Contributed by Peter Mazsa, 30-Jan-2026)

Ref Expression
Assertion dfsuccl4 Suc = { 𝑛 ∣ ∃! 𝑚𝑛 ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) }

Proof

Step Hyp Ref Expression
1 dfsuccl3 Suc = { 𝑛 ∣ ∃! 𝑚 suc 𝑚 = 𝑛 }
2 sucidg ( 𝑚 ∈ V → 𝑚 ∈ suc 𝑚 )
3 2 elv 𝑚 ∈ suc 𝑚
4 eleq2 ( suc 𝑚 = 𝑛 → ( 𝑚 ∈ suc 𝑚𝑚𝑛 ) )
5 3 4 mpbii ( suc 𝑚 = 𝑛𝑚𝑛 )
6 sssucid 𝑚 ⊆ suc 𝑚
7 sseq2 ( suc 𝑚 = 𝑛 → ( 𝑚 ⊆ suc 𝑚𝑚𝑛 ) )
8 6 7 mpbii ( suc 𝑚 = 𝑛𝑚𝑛 )
9 5 8 jca ( suc 𝑚 = 𝑛 → ( 𝑚𝑛𝑚𝑛 ) )
10 9 pm4.71ri ( suc 𝑚 = 𝑛 ↔ ( ( 𝑚𝑛𝑚𝑛 ) ∧ suc 𝑚 = 𝑛 ) )
11 df-3an ( ( 𝑚𝑛𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ↔ ( ( 𝑚𝑛𝑚𝑛 ) ∧ suc 𝑚 = 𝑛 ) )
12 3anass ( ( 𝑚𝑛𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ↔ ( 𝑚𝑛 ∧ ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ) )
13 10 11 12 3bitr2i ( suc 𝑚 = 𝑛 ↔ ( 𝑚𝑛 ∧ ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ) )
14 13 eubii ( ∃! 𝑚 suc 𝑚 = 𝑛 ↔ ∃! 𝑚 ( 𝑚𝑛 ∧ ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ) )
15 df-reu ( ∃! 𝑚𝑛 ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ↔ ∃! 𝑚 ( 𝑚𝑛 ∧ ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) ) )
16 14 15 bitr4i ( ∃! 𝑚 suc 𝑚 = 𝑛 ↔ ∃! 𝑚𝑛 ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) )
17 16 abbii { 𝑛 ∣ ∃! 𝑚 suc 𝑚 = 𝑛 } = { 𝑛 ∣ ∃! 𝑚𝑛 ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) }
18 1 17 eqtri Suc = { 𝑛 ∣ ∃! 𝑚𝑛 ( 𝑚𝑛 ∧ suc 𝑚 = 𝑛 ) }