Metamath Proof Explorer


Theorem madebdaylemold

Description: Lemma for madebday . If the inductive hypothesis of madebday is satisfied, the converse of oldbdayim holds. (Contributed by Scott Fenton, 19-Aug-2024)

Ref Expression
Assertion madebdaylemold ( ( 𝐴 ∈ On ∧ ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( ( bday ‘ 𝑋 ) ∈ 𝐴 → 𝑋 ∈ ( O ‘ 𝐴 ) ) )

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ ( 𝑦 = 𝑋 → ( bday ‘ 𝑦 ) = ( bday ‘ 𝑋 ) )
2 1 sseq1d ⊢ ( 𝑦 = 𝑋 → ( ( bday ‘ 𝑦 ) ⊆ 𝑏 ↔ ( bday ‘ 𝑋 ) ⊆ 𝑏 ) )
3 eleq1 ⊢ ( 𝑦 = 𝑋 → ( 𝑦 ∈ ( M ‘ 𝑏 ) ↔ 𝑋 ∈ ( M ‘ 𝑏 ) ) )
4 2 3 imbi12d ⊢ ( 𝑦 = 𝑋 → ( ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ↔ ( ( bday ‘ 𝑋 ) ⊆ 𝑏 → 𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
5 4 rspcv ⊢ ( 𝑋 ∈ No → ( ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) → ( ( bday ‘ 𝑋 ) ⊆ 𝑏 → 𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
6 5 ralimdv ⊢ ( 𝑋 ∈ No → ( ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) → ∀ 𝑏 ∈ 𝐴 ( ( bday ‘ 𝑋 ) ⊆ 𝑏 → 𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
7 6 impcom ⊢ ( ( ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ∀ 𝑏 ∈ 𝐴 ( ( bday ‘ 𝑋 ) ⊆ 𝑏 → 𝑋 ∈ ( M ‘ 𝑏 ) ) )
8 rexim ⊢ ( ∀ 𝑏 ∈ 𝐴 ( ( bday ‘ 𝑋 ) ⊆ 𝑏 → 𝑋 ∈ ( M ‘ 𝑏 ) ) → ( ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 → ∃ 𝑏 ∈ 𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
9 7 8 syl ⊢ ( ( ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 → ∃ 𝑏 ∈ 𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
10 9 3adant1 ⊢ ( ( 𝐴 ∈ On ∧ ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 → ∃ 𝑏 ∈ 𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
11 bdayon ⊢ ( bday ‘ 𝑋 ) ∈ On
12 onelssex ⊢ ( ( ( bday ‘ 𝑋 ) ∈ On ∧ 𝐴 ∈ On ) → ( ( bday ‘ 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 ) )
13 11 12 mpan ⊢ ( 𝐴 ∈ On → ( ( bday ‘ 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 ) )
14 13 3ad2ant1 ⊢ ( ( 𝐴 ∈ On ∧ ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( ( bday ‘ 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏 ∈ 𝐴 ( bday ‘ 𝑋 ) ⊆ 𝑏 ) )
15 elold ⊢ ( 𝐴 ∈ On → ( 𝑋 ∈ ( O ‘ 𝐴 ) ↔ ∃ 𝑏 ∈ 𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
16 15 3ad2ant1 ⊢ ( ( 𝐴 ∈ On ∧ ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( 𝑋 ∈ ( O ‘ 𝐴 ) ↔ ∃ 𝑏 ∈ 𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
17 10 14 16 3imtr4d ⊢ ( ( 𝐴 ∈ On ∧ ∀ 𝑏 ∈ 𝐴 ∀ 𝑦 ∈ No ( ( bday ‘ 𝑦 ) ⊆ 𝑏 → 𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 ∈ No ) → ( ( bday ‘ 𝑋 ) ∈ 𝐴 → 𝑋 ∈ ( O ‘ 𝐴 ) ) )