Metamath Proof Explorer


Theorem onsucrn

Description: The successor operation is surjective onto its range, the class of successor ordinals. Lemma 1.17 of Schloeder p. 2. (Contributed by RP, 18-Jan-2025)

Ref Expression
Hypothesis onsucrn.f 𝐹 = ( 𝑥 ∈ On ↦ suc 𝑥 )
Assertion onsucrn ran 𝐹 = { 𝑎 ∈ On ∣ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 }

Proof

Step Hyp Ref Expression
1 onsucrn.f 𝐹 = ( 𝑥 ∈ On ↦ suc 𝑥 )
2 simpr ( ( 𝑥 ∈ On ∧ 𝑎 = suc 𝑥 ) → 𝑎 = suc 𝑥 )
3 onsuc ( 𝑥 ∈ On → suc 𝑥 ∈ On )
4 3 adantr ( ( 𝑥 ∈ On ∧ 𝑎 = suc 𝑥 ) → suc 𝑥 ∈ On )
5 2 4 eqeltrd ( ( 𝑥 ∈ On ∧ 𝑎 = suc 𝑥 ) → 𝑎 ∈ On )
6 5 rexlimiva ( ∃ 𝑥 ∈ On 𝑎 = suc 𝑥𝑎 ∈ On )
7 6 pm4.71ri ( ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 ↔ ( 𝑎 ∈ On ∧ ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 ) )
8 suceq ( 𝑥 = 𝑏 → suc 𝑥 = suc 𝑏 )
9 8 eqeq2d ( 𝑥 = 𝑏 → ( 𝑎 = suc 𝑥𝑎 = suc 𝑏 ) )
10 9 cbvrexvw ( ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 ↔ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 )
11 10 anbi2i ( ( 𝑎 ∈ On ∧ ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 ) ↔ ( 𝑎 ∈ On ∧ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 ) )
12 7 11 bitri ( ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 ↔ ( 𝑎 ∈ On ∧ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 ) )
13 12 abbii { 𝑎 ∣ ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 } = { 𝑎 ∣ ( 𝑎 ∈ On ∧ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 ) }
14 1 rnmpt ran 𝐹 = { 𝑎 ∣ ∃ 𝑥 ∈ On 𝑎 = suc 𝑥 }
15 df-rab { 𝑎 ∈ On ∣ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 } = { 𝑎 ∣ ( 𝑎 ∈ On ∧ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 ) }
16 13 14 15 3eqtr4i ran 𝐹 = { 𝑎 ∈ On ∣ ∃ 𝑏 ∈ On 𝑎 = suc 𝑏 }