Theorem csbmpt2 5533

Description: Move substitution into the second part of a maps-to notation. (Contributed by AV, 26-Sep-2019.)

Assertion

Ref	Expression
csbmpt2	⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))

Distinct variable groups:   𝑦,𝐴   𝑦,𝑉   𝑦,𝑌,𝑥

Allowed substitution hints:   𝐴(𝑥)   𝑉(𝑥)   𝑍(𝑥, 𝑦)

Proof of Theorem csbmpt2

Step	Hyp	Ref	Expression
1		csbmpt12 5532	. 2 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))
2		csbconstg 3866	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝑌 = 𝑌)
3	2	mpteq1d 5195	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))
4	1, 3	eqtrd 2796	1 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2145  ⦋csb 3847   ↦ cmpt 5186

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-sn 4585  df-pr 4587  df-op 4591  df-opab 5168  df-mpt 5187

This theorem is used by:  matgsum  22752  csbrdgg  38252