prodeq2si

Description: Equality inference for product. (Contributed by GG, 1-Sep-2025)

		Ref	Expression
	Hypothesis	prodeq2si.1	$⊢ B = C$
	Assertion	prodeq2si	$⊢ \prod_{k \in A} B = \prod_{k \in A} C$

Proof

Step	Hyp	Ref	Expression
1		prodeq2si.1	$⊢ B = C$
2		biid	$⊢ A \subseteq ℤ_{\geq m} \leftrightarrow A \subseteq ℤ_{\geq m}$
3		ifeq1	$⊢ B = C \to if (k \in A, B, 1) = if (k \in A, C, 1)$
4	1 3	ax-mp	$⊢ if (k \in A, B, 1) = if (k \in A, C, 1)$
5	4	mpteq2i	$⊢ (k \in ℤ ⟼ if (k \in A, B, 1)) = (k \in ℤ ⟼ if (k \in A, C, 1))$
6		seqeq3	$⊢ (k \in ℤ ⟼ if (k \in A, B, 1)) = (k \in ℤ ⟼ if (k \in A, C, 1)) \to seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) = seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1)))$
7	5 6	ax-mp	$⊢ seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) = seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1)))$
8	7	breq1i	$⊢ seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y \leftrightarrow seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y$
9	8	anbi2i	$⊢ (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \leftrightarrow (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y)$
10	9	exbii	$⊢ \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \leftrightarrow \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y)$
11	10	rexbii	$⊢ \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \leftrightarrow \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y)$
12		seqeq3	$⊢ (k \in ℤ ⟼ if (k \in A, B, 1)) = (k \in ℤ ⟼ if (k \in A, C, 1)) \to seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) = seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1)))$
13	5 12	ax-mp	$⊢ seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) = seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1)))$
14	13	breq1i	$⊢ seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x \leftrightarrow seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x$
15	2 11 14	3anbi123i	$⊢ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x) \leftrightarrow (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x)$
16	15	rexbii	$⊢ \exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x) \leftrightarrow \exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x)$
17	1	csbeq2i	$⊢ ⦋ f (n) / k ⦌ B = ⦋ f (n) / k ⦌ C$
18	17	mpteq2i	$⊢ (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B) = (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)$
19		seqeq3	$⊢ (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B) = (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C) \to seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C))$
20	18 19	ax-mp	$⊢ seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C))$
21	20	fveq1i	$⊢ seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m) = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m)$
22	21	eqeq2i	$⊢ x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m) \leftrightarrow x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m)$
23	22	anbi2i	$⊢ (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m)) \leftrightarrow (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m))$
24	23	exbii	$⊢ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m)) \leftrightarrow \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m))$
25	24	rexbii	$⊢ \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m)) \leftrightarrow \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m))$
26	16 25	orbi12i	$⊢ (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m))) \leftrightarrow (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m)))$
27	26	iotabii	$⊢ (ι x \| (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m)))) = (ι x \| (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m))))$
28		df-prod	$⊢ \prod_{k \in A} B = (ι x \| (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, B, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ B)) (m))))$
29		df-prod	$⊢ \prod_{k \in A} C = (ι x \| (\exists m \in ℤ (A \subseteq ℤ_{\geq m} \land \exists n \in ℤ_{\geq m} \exists y (y \neq 0 \land seq n (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ y) \land seq m (\times (k \in ℤ ⟼ if (k \in A, C, 1))) ⇝ x) \lor \exists m \in ℕ \exists f (f : (1 \dots m) \underset{1-1 onto}{⟶} A \land x = seq 1 (\times (n \in ℕ ⟼ ⦋ f (n) / k ⦌ C)) (m))))$
30	27 28 29	3eqtr4i	$⊢ \prod_{k \in A} B = \prod_{k \in A} C$

Metamath Proof Explorer

Theorem prodeq2si

Proof