Theorem mpt2eq3dva 5511

Description: Slightly more general equality inference for the maps to notation. (Contributed by NM, 17-Oct-2013.)

Hypothesis

Ref	Expression
mpt2eq3dva.1	⊢ ((φ ∧ x ∈ A ∧ y ∈ B) → 𝐶 = 𝐷)

Assertion

Ref	Expression
mpt2eq3dva	⊢ (φ → (x ∈ A, y ∈ B ↦ 𝐶) = (x ∈ A, y ∈ B ↦ 𝐷))

Distinct variable groups:   φ,x   φ,y

Allowed substitution hints:   A(x,y)   B(x,y)   𝐶(x,y)   𝐷(x,y)

Proof of Theorem mpt2eq3dva

Dummy variable z is distinct from all other variables.

Step	Hyp	Ref	Expression
1		mpt2eq3dva.1	. . . . . 6 ⊢ ((φ ∧ x ∈ A ∧ y ∈ B) → 𝐶 = 𝐷)
2	1	3expb 1104	. . . . 5 ⊢ ((φ ∧ (x ∈ A ∧ y ∈ B)) → 𝐶 = 𝐷)
3	2	eqeq2d 2048	. . . 4 ⊢ ((φ ∧ (x ∈ A ∧ y ∈ B)) → (z = 𝐶 ↔ z = 𝐷))
4	3	pm5.32da 425	. . 3 ⊢ (φ → (((x ∈ A ∧ y ∈ B) ∧ z = 𝐶) ↔ ((x ∈ A ∧ y ∈ B) ∧ z = 𝐷)))
5	4	oprabbidv 5501	. 2 ⊢ (φ → {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = 𝐶)} = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = 𝐷)})
6		df-mpt2 5460	. 2 ⊢ (x ∈ A, y ∈ B ↦ 𝐶) = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = 𝐶)}
7		df-mpt2 5460	. 2 ⊢ (x ∈ A, y ∈ B ↦ 𝐷) = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = 𝐷)}
8	5, 6, 7	3eqtr4g 2094	1 ⊢ (φ → (x ∈ A, y ∈ B ↦ 𝐶) = (x ∈ A, y ∈ B ↦ 𝐷))

Syntax hints:   → wi 4   ∧ wa 97   ∧ w3a 884   = wceq 1242   ∈ wcel 1390  {coprab 5456   ↦ cmpt2 5457

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-11 1394  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-3an 886  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-oprab 5459  df-mpt2 5460

This theorem is referenced by:  mpt2eq3ia  5512  ofeq  5656  fmpt2co  5779  iseqeq2  8895  iseqeq3  8896  iseqval  8900