Theorem csbmpt2 5540

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 5539	. 2 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))
2		csbconstg 3870	. . 3 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌𝑌 = 𝑌)
3	2	mpteq1d 5201	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))
4	1, 3	eqtrd 2798	1 ⊢ (𝐴 ∈ 𝑉 → ⦋𝐴 / 𝑥⦌(𝑦 ∈ 𝑌 ↦ 𝑍) = (𝑦 ∈ 𝑌 ↦ ⦋𝐴 / 𝑥⦌𝑍))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∈ wcel 2147  ⦋csb 3851   ↦ cmpt 5192

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5255  ax-pr 5398

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-sn 4589  df-pr 4591  df-op 4841  df-opab 5174  df-mpt 5193

This theorem is used by:  matgsum  22807  csbrdgg  38520