Metamath Proof Explorer


Theorem sbcoteq1a

Description: Equality theorem for substitution of a class for an ordered triple. (Contributed by Scott Fenton, 22-Aug-2024)

Ref Expression
Assertion sbcoteq1a ⊢ A = x y z → [˙ 1 st ⁡ 1 st ⁡ A / x]˙ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ ↔ φ

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ A = x y z → 2 nd ⁡ A = 2 nd ⁡ x y z
2 ot3rdg ⊢ z ∈ V → 2 nd ⁡ x y z = z
3 2 elv ⊢ 2 nd ⁡ x y z = z
4 1 3 eqtr2di ⊢ A = x y z → z = 2 nd ⁡ A
5 sbceq1a ⊢ z = 2 nd ⁡ A → φ ↔ [˙ 2 nd ⁡ A / z]˙ φ
6 4 5 syl ⊢ A = x y z → φ ↔ [˙ 2 nd ⁡ A / z]˙ φ
7 2fveq3 ⊢ A = x y z → 2 nd ⁡ 1 st ⁡ A = 2 nd ⁡ 1 st ⁡ x y z
8 vex ⊢ x ∈ V
9 vex ⊢ y ∈ V
10 vex ⊢ z ∈ V
11 ot2ndg ⊢ x ∈ V ∧ y ∈ V ∧ z ∈ V → 2 nd ⁡ 1 st ⁡ x y z = y
12 8 9 10 11 mp3an ⊢ 2 nd ⁡ 1 st ⁡ x y z = y
13 7 12 eqtr2di ⊢ A = x y z → y = 2 nd ⁡ 1 st ⁡ A
14 sbceq1a ⊢ y = 2 nd ⁡ 1 st ⁡ A → [˙ 2 nd ⁡ A / z]˙ φ ↔ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ
15 13 14 syl ⊢ A = x y z → [˙ 2 nd ⁡ A / z]˙ φ ↔ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ
16 2fveq3 ⊢ A = x y z → 1 st ⁡ 1 st ⁡ A = 1 st ⁡ 1 st ⁡ x y z
17 ot1stg ⊢ x ∈ V ∧ y ∈ V ∧ z ∈ V → 1 st ⁡ 1 st ⁡ x y z = x
18 8 9 10 17 mp3an ⊢ 1 st ⁡ 1 st ⁡ x y z = x
19 16 18 eqtr2di ⊢ A = x y z → x = 1 st ⁡ 1 st ⁡ A
20 sbceq1a ⊢ x = 1 st ⁡ 1 st ⁡ A → [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ ↔ [˙ 1 st ⁡ 1 st ⁡ A / x]˙ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ
21 19 20 syl ⊢ A = x y z → [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ ↔ [˙ 1 st ⁡ 1 st ⁡ A / x]˙ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ
22 6 15 21 3bitrrd ⊢ A = x y z → [˙ 1 st ⁡ 1 st ⁡ A / x]˙ [˙ 2 nd ⁡ 1 st ⁡ A / y]˙ [˙ 2 nd ⁡ A / z]˙ φ ↔ φ