Metamath Proof Explorer


Theorem csbnestgf

Description: Nest the composition of two substitutions. Usage of this theorem is discouraged because it depends on ax-13 . Use the weaker csbnestgfw when possible. (Contributed by NM, 23-Nov-2005) (Proof shortened by Mario Carneiro, 10-Nov-2016) (New usage is discouraged.)

Ref Expression
Assertion csbnestgf ( ( 𝐴𝑉 ∧ ∀ 𝑦 𝑥 𝐶 ) → 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 = 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 )

Proof

Step Hyp Ref Expression
1 elex ( 𝐴𝑉𝐴 ∈ V )
2 df-csb 𝐵 / 𝑦 𝐶 = { 𝑧[ 𝐵 / 𝑦 ] 𝑧𝐶 }
3 2 abeq2i ( 𝑧 𝐵 / 𝑦 𝐶[ 𝐵 / 𝑦 ] 𝑧𝐶 )
4 3 sbcbii ( [ 𝐴 / 𝑥 ] 𝑧 𝐵 / 𝑦 𝐶[ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝑧𝐶 )
5 nfcr ( 𝑥 𝐶 → Ⅎ 𝑥 𝑧𝐶 )
6 5 alimi ( ∀ 𝑦 𝑥 𝐶 → ∀ 𝑦𝑥 𝑧𝐶 )
7 sbcnestgf ( ( 𝐴 ∈ V ∧ ∀ 𝑦𝑥 𝑧𝐶 ) → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝑧𝐶[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 ) )
8 6 7 sylan2 ( ( 𝐴 ∈ V ∧ ∀ 𝑦 𝑥 𝐶 ) → ( [ 𝐴 / 𝑥 ] [ 𝐵 / 𝑦 ] 𝑧𝐶[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 ) )
9 4 8 bitrid ( ( 𝐴 ∈ V ∧ ∀ 𝑦 𝑥 𝐶 ) → ( [ 𝐴 / 𝑥 ] 𝑧 𝐵 / 𝑦 𝐶[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 ) )
10 9 abbidv ( ( 𝐴 ∈ V ∧ ∀ 𝑦 𝑥 𝐶 ) → { 𝑧[ 𝐴 / 𝑥 ] 𝑧 𝐵 / 𝑦 𝐶 } = { 𝑧[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 } )
11 1 10 sylan ( ( 𝐴𝑉 ∧ ∀ 𝑦 𝑥 𝐶 ) → { 𝑧[ 𝐴 / 𝑥 ] 𝑧 𝐵 / 𝑦 𝐶 } = { 𝑧[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 } )
12 df-csb 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 = { 𝑧[ 𝐴 / 𝑥 ] 𝑧 𝐵 / 𝑦 𝐶 }
13 df-csb 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 = { 𝑧[ 𝐴 / 𝑥 𝐵 / 𝑦 ] 𝑧𝐶 }
14 11 12 13 3eqtr4g ( ( 𝐴𝑉 ∧ ∀ 𝑦 𝑥 𝐶 ) → 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 = 𝐴 / 𝑥 𝐵 / 𝑦 𝐶 )