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 5194	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))
4	1, 3	eqtrd 2795	1 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ∈ wcel 2144  ⦋csb 3847   ↦ cmpt 5185

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-rab 3413  df-v 3452  df-sbc 3739  df-csb 3848  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-sn 4585  df-pr 4587  df-op 4835  df-opab 5167  df-mpt 5186

This theorem is used by:  matgsum  22761  csbrdgg  38389