Theorem curry2val

Description: The value of a curried function with a constant second argument. (Contributed by NM, 16-Dec-2008)

	Ref	Expression
Hypothesis	curry2.1	⊢ 𝐺 = ( 𝐹 ∘ ◡ ( 1st ↾ ( V × { 𝐶 } ) ) )
Assertion	curry2val	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → ( 𝐺 ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		curry2.1	⊢ 𝐺 = ( 𝐹 ∘ ◡ ( 1st ↾ ( V × { 𝐶 } ) ) )
2	1	curry2	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → 𝐺 = ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) )
3	2	fveq1d	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → ( 𝐺 ‘ 𝐷 ) = ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) )
4		eqid	⊢ ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) = ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) )
5	4	fvmptndm	⊢ ( ¬ 𝐷 ∈ 𝐴 → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ∅ )
6	5	adantl	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ ¬ 𝐷 ∈ 𝐴 ) → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ∅ )
7		fndm	⊢ ( 𝐹 Fn ( 𝐴 × 𝐵 ) → dom 𝐹 = ( 𝐴 × 𝐵 ) )
8		simpl	⊢ ( ( 𝐷 ∈ 𝐴 ∧ 𝐶 ∈ 𝐵 ) → 𝐷 ∈ 𝐴 )
9	8	con3i	⊢ ( ¬ 𝐷 ∈ 𝐴 → ¬ ( 𝐷 ∈ 𝐴 ∧ 𝐶 ∈ 𝐵 ) )
10		ndmovg	⊢ ( ( dom 𝐹 = ( 𝐴 × 𝐵 ) ∧ ¬ ( 𝐷 ∈ 𝐴 ∧ 𝐶 ∈ 𝐵 ) ) → ( 𝐷 𝐹 𝐶 ) = ∅ )
11	7 9 10	syl2an	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ ¬ 𝐷 ∈ 𝐴 ) → ( 𝐷 𝐹 𝐶 ) = ∅ )
12	6 11	eqtr4d	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ ¬ 𝐷 ∈ 𝐴 ) → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) )
13	12	ex	⊢ ( 𝐹 Fn ( 𝐴 × 𝐵 ) → ( ¬ 𝐷 ∈ 𝐴 → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) ) )
14	13	adantr	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → ( ¬ 𝐷 ∈ 𝐴 → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) ) )
15		oveq1	⊢ ( 𝑥 = 𝐷 → ( 𝑥 𝐹 𝐶 ) = ( 𝐷 𝐹 𝐶 ) )
16		ovex	⊢ ( 𝐷 𝐹 𝐶 ) ∈ V
17	15 4 16	fvmpt	⊢ ( 𝐷 ∈ 𝐴 → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) )
18	14 17	pm2.61d2	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → ( ( 𝑥 ∈ 𝐴 ↦ ( 𝑥 𝐹 𝐶 ) ) ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) )
19	3 18	eqtrd	⊢ ( ( 𝐹 Fn ( 𝐴 × 𝐵 ) ∧ 𝐶 ∈ 𝐵 ) → ( 𝐺 ‘ 𝐷 ) = ( 𝐷 𝐹 𝐶 ) )