Metamath Proof Explorer


Theorem eloprabi

Description: A consequence of membership in an operation class abstraction, using ordered pair extractors. (Contributed by NM, 6-Nov-2006) (Revised by David Abernethy, 19-Jun-2012)

Ref Expression
Hypotheses eloprabi.1 ⊢ x = 1 st ⁡ 1 st ⁡ A → φ ↔ ψ
eloprabi.2 ⊢ y = 2 nd ⁡ 1 st ⁡ A → ψ ↔ χ
eloprabi.3 ⊢ z = 2 nd ⁡ A → χ ↔ θ
Assertion eloprabi ⊢ A ∈ x y z | φ → θ

Proof

Step Hyp Ref Expression
1 eloprabi.1 ⊢ x = 1 st ⁡ 1 st ⁡ A → φ ↔ ψ
2 eloprabi.2 ⊢ y = 2 nd ⁡ 1 st ⁡ A → ψ ↔ χ
3 eloprabi.3 ⊢ z = 2 nd ⁡ A → χ ↔ θ
4 eqeq1 ⊢ w = A → w = x y z ↔ A = x y z
5 4 anbi1d ⊢ w = A → w = x y z ∧ φ ↔ A = x y z ∧ φ
6 5 3exbidv ⊢ w = A → ∃ x ∃ y ∃ z w = x y z ∧ φ ↔ ∃ x ∃ y ∃ z A = x y z ∧ φ
7 df-oprab ⊢ x y z | φ = w | ∃ x ∃ y ∃ z w = x y z ∧ φ
8 6 7 elab2g ⊢ A ∈ x y z | φ → A ∈ x y z | φ ↔ ∃ x ∃ y ∃ z A = x y z ∧ φ
9 8 ibi ⊢ A ∈ x y z | φ → ∃ x ∃ y ∃ z A = x y z ∧ φ
10 opex ⊢ x y ∈ V
11 vex ⊢ z ∈ V
12 10 11 op1std ⊢ A = x y z → 1 st ⁡ A = x y
13 12 fveq2d ⊢ A = x y z → 1 st ⁡ 1 st ⁡ A = 1 st ⁡ x y
14 vex ⊢ x ∈ V
15 vex ⊢ y ∈ V
16 14 15 op1st ⊢ 1 st ⁡ x y = x
17 13 16 eqtr2di ⊢ A = x y z → x = 1 st ⁡ 1 st ⁡ A
18 17 1 syl ⊢ A = x y z → φ ↔ ψ
19 12 fveq2d ⊢ A = x y z → 2 nd ⁡ 1 st ⁡ A = 2 nd ⁡ x y
20 14 15 op2nd ⊢ 2 nd ⁡ x y = y
21 19 20 eqtr2di ⊢ A = x y z → y = 2 nd ⁡ 1 st ⁡ A
22 21 2 syl ⊢ A = x y z → ψ ↔ χ
23 10 11 op2ndd ⊢ A = x y z → 2 nd ⁡ A = z
24 23 eqcomd ⊢ A = x y z → z = 2 nd ⁡ A
25 24 3 syl ⊢ A = x y z → χ ↔ θ
26 18 22 25 3bitrd ⊢ A = x y z → φ ↔ θ
27 26 biimpa ⊢ A = x y z ∧ φ → θ
28 27 exlimiv ⊢ ∃ z A = x y z ∧ φ → θ
29 28 exlimiv ⊢ ∃ y ∃ z A = x y z ∧ φ → θ
30 29 exlimiv ⊢ ∃ x ∃ y ∃ z A = x y z ∧ φ → θ
31 9 30 syl ⊢ A ∈ x y z | φ → θ