Metamath Proof Explorer


Theorem f1od

Description: Describe an implicit one-to-one onto function. (Contributed by Mario Carneiro, 12-May-2014)

Ref Expression
Hypotheses f1od.1 ⊢ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝐶 )
f1od.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐶 ∈ 𝑊 )
f1od.3 ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐵 ) → 𝐷 ∈ 𝑋 )
f1od.4 ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝑥 = 𝐷 ) ) )
Assertion f1od ( 𝜑 → 𝐹 : 𝐴 –1-1-onto→ 𝐵 )

Proof

Step Hyp Ref Expression
1 f1od.1 ⊢ 𝐹 = ( 𝑥 ∈ 𝐴 ↦ 𝐶 )
2 f1od.2 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → 𝐶 ∈ 𝑊 )
3 f1od.3 ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐵 ) → 𝐷 ∈ 𝑋 )
4 f1od.4 ⊢ ( 𝜑 → ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 = 𝐶 ) ↔ ( 𝑦 ∈ 𝐵 ∧ 𝑥 = 𝐷 ) ) )
5 1 2 3 4 f1ocnvd ⊢ ( 𝜑 → ( 𝐹 : 𝐴 –1-1-onto→ 𝐵 ∧ ◡ 𝐹 = ( 𝑦 ∈ 𝐵 ↦ 𝐷 ) ) )
6 5 simpld ⊢ ( 𝜑 → 𝐹 : 𝐴 –1-1-onto→ 𝐵 )