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Metamath Proof Explorer

Theorem f1cofveqaeq

Description: If the values of a composition of one-to-one functions for two arguments are equal, the arguments themselves must be equal. (Contributed by AV, 3-Feb-2021)

Ref Expression
Assertion f1cofveqaeq ( ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐹 ‘ ( 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ 𝑌 ) ) → 𝑋 = 𝑌 ) )

Proof

Step Hyp Ref Expression
1 simpl ⊢ ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) → 𝐹 : 𝐵 –1-1→ 𝐶 )
2 f1f ⊢ ( 𝐺 : 𝐴 –1-1→ 𝐵 → 𝐺 : 𝐴 ⟶ 𝐵 )
3 ffvelcdm ⊢ ( ( 𝐺 : 𝐴 ⟶ 𝐵 ∧ 𝑋 ∈ 𝐴 ) → ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 )
4 3 ex ⊢ ( 𝐺 : 𝐴 ⟶ 𝐵 → ( 𝑋 ∈ 𝐴 → ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ) )
5 ffvelcdm ⊢ ( ( 𝐺 : 𝐴 ⟶ 𝐵 ∧ 𝑌 ∈ 𝐴 ) → ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 )
6 5 ex ⊢ ( 𝐺 : 𝐴 ⟶ 𝐵 → ( 𝑌 ∈ 𝐴 → ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) )
7 4 6 anim12d ⊢ ( 𝐺 : 𝐴 ⟶ 𝐵 → ( ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) → ( ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ∧ ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) ) )
8 2 7 syl ⊢ ( 𝐺 : 𝐴 –1-1→ 𝐵 → ( ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) → ( ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ∧ ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) ) )
9 8 adantl ⊢ ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) → ( ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) → ( ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ∧ ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) ) )
10 9 imp ⊢ ( ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ∧ ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) )
11 f1veqaeq ⊢ ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ ( ( 𝐺 ‘ 𝑋 ) ∈ 𝐵 ∧ ( 𝐺 ‘ 𝑌 ) ∈ 𝐵 ) ) → ( ( 𝐹 ‘ ( 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ 𝑌 ) ) → ( 𝐺 ‘ 𝑋 ) = ( 𝐺 ‘ 𝑌 ) ) )
12 1 10 11 syl2an2r ⊢ ( ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐹 ‘ ( 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ 𝑌 ) ) → ( 𝐺 ‘ 𝑋 ) = ( 𝐺 ‘ 𝑌 ) ) )
13 f1veqaeq ⊢ ( ( 𝐺 : 𝐴 –1-1→ 𝐵 ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐺 ‘ 𝑋 ) = ( 𝐺 ‘ 𝑌 ) → 𝑋 = 𝑌 ) )
14 13 adantll ⊢ ( ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐺 ‘ 𝑋 ) = ( 𝐺 ‘ 𝑌 ) → 𝑋 = 𝑌 ) )
15 12 14 syld ⊢ ( ( ( 𝐹 : 𝐵 –1-1→ 𝐶 ∧ 𝐺 : 𝐴 –1-1→ 𝐵 ) ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ) ) → ( ( 𝐹 ‘ ( 𝐺 ‘ 𝑋 ) ) = ( 𝐹 ‘ ( 𝐺 ‘ 𝑌 ) ) → 𝑋 = 𝑌 ) )