# Metamath Proof Explorer

## Theorem eqfnfv2f

Description: Equality of functions is determined by their values. Special case of Exercise 4 of TakeutiZaring p. 28 (with domain equality omitted). This version of eqfnfv uses bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 29-Jan-2004)

Ref Expression
Hypotheses eqfnfv2f.1 𝑥 𝐹
eqfnfv2f.2 𝑥 𝐺
Assertion eqfnfv2f ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )

### Proof

Step Hyp Ref Expression
1 eqfnfv2f.1 𝑥 𝐹
2 eqfnfv2f.2 𝑥 𝐺
3 eqfnfv ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ) )
4 nfcv 𝑥 𝑧
5 1 4 nffv 𝑥 ( 𝐹𝑧 )
6 2 4 nffv 𝑥 ( 𝐺𝑧 )
7 5 6 nfeq 𝑥 ( 𝐹𝑧 ) = ( 𝐺𝑧 )
8 nfv 𝑧 ( 𝐹𝑥 ) = ( 𝐺𝑥 )
9 fveq2 ( 𝑧 = 𝑥 → ( 𝐹𝑧 ) = ( 𝐹𝑥 ) )
10 fveq2 ( 𝑧 = 𝑥 → ( 𝐺𝑧 ) = ( 𝐺𝑥 ) )
11 9 10 eqeq12d ( 𝑧 = 𝑥 → ( ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ↔ ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
12 7 8 11 cbvralw ( ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
13 3 12 syl6bb ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
The text passage is a formal proof in the Metamath Proof Explorer, which deals with the equality of functions based on their values. It does not contain any mathematical questions or answers in the form of word problems or calculations.
no_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Metamath Proof Explorer

## Theorem eqfnfv2f

Description: Equality of functions is determined by their values. Special case of Exercise 4 of TakeutiZaring p. 28 (with domain equality omitted). This version of eqfnfv uses bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 29-Jan-2004)

Ref Expression
Hypotheses eqfnfv2f.1 𝑥 𝐹
eqfnfv2f.2 𝑥 𝐺
Assertion eqfnfv2f ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )

### Proof

Step Hyp Ref Expression
1 eqfnfv2f.1 𝑥 𝐹
2 eqfnfv2f.2 𝑥 𝐺
3 eqfnfv ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ) )
4 nfcv 𝑥 𝑧
5 1 4 nffv 𝑥 ( 𝐹𝑧 )
6 2 4 nffv 𝑥 ( 𝐺𝑧 )
7 5 6 nfeq 𝑥 ( 𝐹𝑧 ) = ( 𝐺𝑧 )
8 nfv 𝑧 ( 𝐹𝑥 ) = ( 𝐺𝑥 )
9 fveq2 ( 𝑧 = 𝑥 → ( 𝐹𝑧 ) = ( 𝐹𝑥 ) )
10 fveq2 ( 𝑧 = 𝑥 → ( 𝐺𝑧 ) = ( 𝐺𝑥 ) )
11 9 10 eqeq12d ( 𝑧 = 𝑥 → ( ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ↔ ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
12 7 8 11 cbvralw ( ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝐺𝑧 ) ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) )
13 3 12 syl6bb ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥𝐴 ( 𝐹𝑥 ) = ( 𝐺𝑥 ) ) )
user
The text passage is a formal proof in the Metamath Proof Explorer, which deals with the equality of functions based on their values. It does not contain any mathematical questions or answers in the form of word problems or calculations.

TYPE: NO_QUESTION