Metamath Proof Explorer


Theorem unblem1

Description: Lemma for unbnn . After removing the successor of an element from an unbounded set of natural numbers, the intersection of the result belongs to the original unbounded set. (Contributed by NM, 3-Dec-2003)

Ref Expression
Assertion unblem1 ( ( ( 𝐵 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ) ∧ 𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ∈ 𝐵 )

Proof

Step Hyp Ref Expression
1 omsson ω ⊆ On
2 sstr ( ( 𝐵 ⊆ ω ∧ ω ⊆ On ) → 𝐵 ⊆ On )
3 1 2 mpan2 ( 𝐵 ⊆ ω → 𝐵 ⊆ On )
4 3 ssdifssd ( 𝐵 ⊆ ω → ( 𝐵 ∖ suc 𝐴 ) ⊆ On )
5 4 ad2antrr ( ( ( 𝐵 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ) ∧ 𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ⊆ On )
6 ssel ( 𝐵 ⊆ ω → ( 𝐴𝐵𝐴 ∈ ω ) )
7 peano2b ( 𝐴 ∈ ω ↔ suc 𝐴 ∈ ω )
8 6 7 syl6ib ( 𝐵 ⊆ ω → ( 𝐴𝐵 → suc 𝐴 ∈ ω ) )
9 eleq1 ( 𝑥 = suc 𝐴 → ( 𝑥𝑦 ↔ suc 𝐴𝑦 ) )
10 9 rexbidv ( 𝑥 = suc 𝐴 → ( ∃ 𝑦𝐵 𝑥𝑦 ↔ ∃ 𝑦𝐵 suc 𝐴𝑦 ) )
11 10 rspccva ( ( ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ∧ suc 𝐴 ∈ ω ) → ∃ 𝑦𝐵 suc 𝐴𝑦 )
12 ssel ( 𝐵 ⊆ ω → ( 𝑦𝐵𝑦 ∈ ω ) )
13 nnord ( 𝑦 ∈ ω → Ord 𝑦 )
14 ordn2lp ( Ord 𝑦 → ¬ ( 𝑦 ∈ suc 𝐴 ∧ suc 𝐴𝑦 ) )
15 imnan ( ( 𝑦 ∈ suc 𝐴 → ¬ suc 𝐴𝑦 ) ↔ ¬ ( 𝑦 ∈ suc 𝐴 ∧ suc 𝐴𝑦 ) )
16 14 15 sylibr ( Ord 𝑦 → ( 𝑦 ∈ suc 𝐴 → ¬ suc 𝐴𝑦 ) )
17 16 con2d ( Ord 𝑦 → ( suc 𝐴𝑦 → ¬ 𝑦 ∈ suc 𝐴 ) )
18 13 17 syl ( 𝑦 ∈ ω → ( suc 𝐴𝑦 → ¬ 𝑦 ∈ suc 𝐴 ) )
19 12 18 syl6 ( 𝐵 ⊆ ω → ( 𝑦𝐵 → ( suc 𝐴𝑦 → ¬ 𝑦 ∈ suc 𝐴 ) ) )
20 19 imdistand ( 𝐵 ⊆ ω → ( ( 𝑦𝐵 ∧ suc 𝐴𝑦 ) → ( 𝑦𝐵 ∧ ¬ 𝑦 ∈ suc 𝐴 ) ) )
21 eldif ( 𝑦 ∈ ( 𝐵 ∖ suc 𝐴 ) ↔ ( 𝑦𝐵 ∧ ¬ 𝑦 ∈ suc 𝐴 ) )
22 ne0i ( 𝑦 ∈ ( 𝐵 ∖ suc 𝐴 ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ )
23 21 22 sylbir ( ( 𝑦𝐵 ∧ ¬ 𝑦 ∈ suc 𝐴 ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ )
24 20 23 syl6 ( 𝐵 ⊆ ω → ( ( 𝑦𝐵 ∧ suc 𝐴𝑦 ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) )
25 24 expd ( 𝐵 ⊆ ω → ( 𝑦𝐵 → ( suc 𝐴𝑦 → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) ) )
26 25 rexlimdv ( 𝐵 ⊆ ω → ( ∃ 𝑦𝐵 suc 𝐴𝑦 → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) )
27 11 26 syl5 ( 𝐵 ⊆ ω → ( ( ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ∧ suc 𝐴 ∈ ω ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) )
28 8 27 sylan2d ( 𝐵 ⊆ ω → ( ( ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) )
29 28 impl ( ( ( 𝐵 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ) ∧ 𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ )
30 onint ( ( ( 𝐵 ∖ suc 𝐴 ) ⊆ On ∧ ( 𝐵 ∖ suc 𝐴 ) ≠ ∅ ) → ( 𝐵 ∖ suc 𝐴 ) ∈ ( 𝐵 ∖ suc 𝐴 ) )
31 5 29 30 syl2anc ( ( ( 𝐵 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ) ∧ 𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ∈ ( 𝐵 ∖ suc 𝐴 ) )
32 31 eldifad ( ( ( 𝐵 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦𝐵 𝑥𝑦 ) ∧ 𝐴𝐵 ) → ( 𝐵 ∖ suc 𝐴 ) ∈ 𝐵 )