Metamath Proof Explorer


Theorem csbwrecsg

Description: Move class substitution in and out of the well-founded recursive function generator. (Contributed by ML, 25-Oct-2020) (Revised by Scott Fenton, 18-Nov-2024)

Ref Expression
Assertion csbwrecsg ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ wrecs ( 𝑅 , 𝐷 , 𝐹 ) = wrecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ) )

Proof

Step Hyp Ref Expression
1 csbfrecsg ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ frecs ( 𝑅 , 𝐷 , ( 𝐹 ∘ 2nd ) ) = frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) ) )
2 eqid ⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝑅 = ⦋ 𝐴 / 𝑥 ⦌ 𝑅
3 eqid ⊢ ⦋ 𝐴 / 𝑥 ⦌ 𝐷 = ⦋ 𝐴 / 𝑥 ⦌ 𝐷
4 csbcog ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ ⦋ 𝐴 / 𝑥 ⦌ 2nd ) )
5 csbconstg ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ 2nd = 2nd )
6 5 coeq2d ⊢ ( 𝐴 ∈ 𝑉 → ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ ⦋ 𝐴 / 𝑥 ⦌ 2nd ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) )
7 4 6 eqtrd ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) )
8 frecseq123 ⊢ ( ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 = ⦋ 𝐴 / 𝑥 ⦌ 𝑅 ∧ ⦋ 𝐴 / 𝑥 ⦌ 𝐷 = ⦋ 𝐴 / 𝑥 ⦌ 𝐷 ∧ ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) = ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) ) → frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) ) = frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) ) )
9 2 3 7 8 mp3an12i ⊢ ( 𝐴 ∈ 𝑉 → frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ ( 𝐹 ∘ 2nd ) ) = frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) ) )
10 1 9 eqtrd ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ frecs ( 𝑅 , 𝐷 , ( 𝐹 ∘ 2nd ) ) = frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) ) )
11 df-wrecs ⊢ wrecs ( 𝑅 , 𝐷 , 𝐹 ) = frecs ( 𝑅 , 𝐷 , ( 𝐹 ∘ 2nd ) )
12 11 csbeq2i ⊢ ⦋ 𝐴 / 𝑥 ⦌ wrecs ( 𝑅 , 𝐷 , 𝐹 ) = ⦋ 𝐴 / 𝑥 ⦌ frecs ( 𝑅 , 𝐷 , ( 𝐹 ∘ 2nd ) )
13 df-wrecs ⊢ wrecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ) = frecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ( ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ∘ 2nd ) )
14 10 12 13 3eqtr4g ⊢ ( 𝐴 ∈ 𝑉 → ⦋ 𝐴 / 𝑥 ⦌ wrecs ( 𝑅 , 𝐷 , 𝐹 ) = wrecs ( ⦋ 𝐴 / 𝑥 ⦌ 𝑅 , ⦋ 𝐴 / 𝑥 ⦌ 𝐷 , ⦋ 𝐴 / 𝑥 ⦌ 𝐹 ) )