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 ∧ ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( ( bday 𝑋 ) ∈ 𝐴𝑋 ∈ ( 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 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ↔ ( ( bday 𝑋 ) ⊆ 𝑏𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
5 4 rspcv ( 𝑋 No → ( ∀ 𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) → ( ( bday 𝑋 ) ⊆ 𝑏𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
6 5 ralimdv ( 𝑋 No → ( ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) → ∀ 𝑏𝐴 ( ( bday 𝑋 ) ⊆ 𝑏𝑋 ∈ ( M ‘ 𝑏 ) ) ) )
7 6 impcom ( ( ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ∀ 𝑏𝐴 ( ( bday 𝑋 ) ⊆ 𝑏𝑋 ∈ ( M ‘ 𝑏 ) ) )
8 rexim ( ∀ 𝑏𝐴 ( ( bday 𝑋 ) ⊆ 𝑏𝑋 ∈ ( M ‘ 𝑏 ) ) → ( ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 → ∃ 𝑏𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
9 7 8 syl ( ( ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 → ∃ 𝑏𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
10 9 3adant1 ( ( 𝐴 ∈ On ∧ ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 → ∃ 𝑏𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
11 bdayelon ( bday 𝑋 ) ∈ On
12 onelssex ( ( ( bday 𝑋 ) ∈ On ∧ 𝐴 ∈ On ) → ( ( bday 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 ) )
13 11 12 mpan ( 𝐴 ∈ On → ( ( bday 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 ) )
14 13 3ad2ant1 ( ( 𝐴 ∈ On ∧ ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( ( bday 𝑋 ) ∈ 𝐴 ↔ ∃ 𝑏𝐴 ( bday 𝑋 ) ⊆ 𝑏 ) )
15 elold ( 𝐴 ∈ On → ( 𝑋 ∈ ( O ‘ 𝐴 ) ↔ ∃ 𝑏𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
16 15 3ad2ant1 ( ( 𝐴 ∈ On ∧ ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( 𝑋 ∈ ( O ‘ 𝐴 ) ↔ ∃ 𝑏𝐴 𝑋 ∈ ( M ‘ 𝑏 ) ) )
17 10 14 16 3imtr4d ( ( 𝐴 ∈ On ∧ ∀ 𝑏𝐴𝑦 No ( ( bday 𝑦 ) ⊆ 𝑏𝑦 ∈ ( M ‘ 𝑏 ) ) ∧ 𝑋 No ) → ( ( bday 𝑋 ) ∈ 𝐴𝑋 ∈ ( O ‘ 𝐴 ) ) )