Theorem mpt2eq3dva 5670

Description: Slightly more general equality inference for the maps to notation. (Contributed by set.mm contributors, 17-Oct-2013.) (Revised by set.mm contributors, 16-Dec-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 1152	. . . . 5 ⊢ ((φ ∧ (x ∈ A ∧ y ∈ B)) → C = D)
3	2	eqeq2d 2364	. . . 4 ⊢ ((φ ∧ (x ∈ A ∧ y ∈ B)) → (z = C ↔ z = D))
4	3	pm5.32da 622	. . 3 ⊢ (φ → (((x ∈ A ∧ y ∈ B) ∧ z = C) ↔ ((x ∈ A ∧ y ∈ B) ∧ z = D)))
5	4	oprabbidv 5565	. 2 ⊢ (φ → {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = C)} = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = D)})
6		df-mpt2 5655	. 2 ⊢ (x ∈ A, y ∈ B ↦ C) = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = C)}
7		df-mpt2 5655	. 2 ⊢ (x ∈ A, y ∈ B ↦ D) = {⟨⟨x, y⟩, z⟩ ∣ ((x ∈ A ∧ y ∈ B) ∧ z = D)}
8	5, 6, 7	3eqtr4g 2410	1 ⊢ (φ → (x ∈ A, y ∈ B ↦ C) = (x ∈ A, y ∈ B ↦ D))

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   ∧ w3a 934   = wceq 1642   ∈ wcel 1710  {coprab 5528   ↦ cmpt2 5654

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-oprab 5529  df-mpt2 5655

This theorem is used by:  mpt2eq3ia  5671