Metamath Proof Explorer


Theorem f1ocan1fv

Description: Cancel a composition by a bijection by preapplying the converse. (Contributed by Jeff Madsen, 2-Sep-2009) (Proof shortened by Mario Carneiro, 27-Dec-2014)

Ref Expression
Assertion f1ocan1fv ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝐹 ∘ 𝐺 ) ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ 𝑋 ) )

Proof

Step Hyp Ref Expression
1 f1of ⊢ ( 𝐺 : 𝐴 –1-1-onto→ 𝐵 → 𝐺 : 𝐴 ⟶ 𝐵 )
2 1 3ad2ant2 ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → 𝐺 : 𝐴 ⟶ 𝐵 )
3 f1ocnv ⊢ ( 𝐺 : 𝐴 –1-1-onto→ 𝐵 → ◡ 𝐺 : 𝐵 –1-1-onto→ 𝐴 )
4 f1of ⊢ ( ◡ 𝐺 : 𝐵 –1-1-onto→ 𝐴 → ◡ 𝐺 : 𝐵 ⟶ 𝐴 )
5 3 4 syl ⊢ ( 𝐺 : 𝐴 –1-1-onto→ 𝐵 → ◡ 𝐺 : 𝐵 ⟶ 𝐴 )
6 5 3ad2ant2 ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ◡ 𝐺 : 𝐵 ⟶ 𝐴 )
7 simp3 ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 )
8 6 7 ffvelcdmd ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( ◡ 𝐺 ‘ 𝑋 ) ∈ 𝐴 )
9 fvco3 ⊢ ( ( 𝐺 : 𝐴 ⟶ 𝐵 ∧ ( ◡ 𝐺 ‘ 𝑋 ) ∈ 𝐴 ) → ( ( 𝐹 ∘ 𝐺 ) ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) ) )
10 2 8 9 syl2anc ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝐹 ∘ 𝐺 ) ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) ) )
11 f1ocnvfv2 ⊢ ( ( 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( 𝐺 ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = 𝑋 )
12 11 3adant1 ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( 𝐺 ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = 𝑋 )
13 12 fveq2d ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( 𝐹 ‘ ( 𝐺 ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) ) = ( 𝐹 ‘ 𝑋 ) )
14 10 13 eqtrd ⊢ ( ( Fun 𝐹 ∧ 𝐺 : 𝐴 –1-1-onto→ 𝐵 ∧ 𝑋 ∈ 𝐵 ) → ( ( 𝐹 ∘ 𝐺 ) ‘ ( ◡ 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ 𝑋 ) )