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Metamath Proof Explorer

Theorem fthpropd

Description: If two categories have the same set of objects, morphisms, and compositions, then they have the same faithful functors. (Contributed by Mario Carneiro, 27-Jan-2017)

Ref Expression
Hypotheses fullpropd.1 ⊢ ( 𝜑 → ( Homf ‘ 𝐴 ) = ( Homf ‘ 𝐵 ) )
fullpropd.2 ⊢ ( 𝜑 → ( compf ‘ 𝐴 ) = ( compf ‘ 𝐵 ) )
fullpropd.3 ⊢ ( 𝜑 → ( Homf ‘ 𝐶 ) = ( Homf ‘ 𝐷 ) )
fullpropd.4 ⊢ ( 𝜑 → ( compf ‘ 𝐶 ) = ( compf ‘ 𝐷 ) )
fullpropd.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
fullpropd.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑉 )
fullpropd.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑉 )
fullpropd.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑉 )
Assertion fthpropd ( 𝜑 → ( 𝐴 Faith 𝐶 ) = ( 𝐵 Faith 𝐷 ) )

Proof

Step Hyp Ref Expression
1 fullpropd.1 ⊢ ( 𝜑 → ( Homf ‘ 𝐴 ) = ( Homf ‘ 𝐵 ) )
2 fullpropd.2 ⊢ ( 𝜑 → ( compf ‘ 𝐴 ) = ( compf ‘ 𝐵 ) )
3 fullpropd.3 ⊢ ( 𝜑 → ( Homf ‘ 𝐶 ) = ( Homf ‘ 𝐷 ) )
4 fullpropd.4 ⊢ ( 𝜑 → ( compf ‘ 𝐶 ) = ( compf ‘ 𝐷 ) )
5 fullpropd.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
6 fullpropd.b ⊢ ( 𝜑 → 𝐵 ∈ 𝑉 )
7 fullpropd.c ⊢ ( 𝜑 → 𝐶 ∈ 𝑉 )
8 fullpropd.d ⊢ ( 𝜑 → 𝐷 ∈ 𝑉 )
9 relfth ⊢ Rel ( 𝐴 Faith 𝐶 )
10 relfth ⊢ Rel ( 𝐵 Faith 𝐷 )
11 1 2 3 4 5 6 7 8 funcpropd ⊢ ( 𝜑 → ( 𝐴 Func 𝐶 ) = ( 𝐵 Func 𝐷 ) )
12 11 breqd ⊢ ( 𝜑 → ( 𝑓 ( 𝐴 Func 𝐶 ) 𝑔 ↔ 𝑓 ( 𝐵 Func 𝐷 ) 𝑔 ) )
13 1 homfeqbas ⊢ ( 𝜑 → ( Base ‘ 𝐴 ) = ( Base ‘ 𝐵 ) )
14 13 raleqdv ⊢ ( 𝜑 → ( ∀ 𝑦 ∈ ( Base ‘ 𝐴 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ↔ ∀ 𝑦 ∈ ( Base ‘ 𝐵 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) )
15 13 14 raleqbidv ⊢ ( 𝜑 → ( ∀ 𝑥 ∈ ( Base ‘ 𝐴 ) ∀ 𝑦 ∈ ( Base ‘ 𝐴 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ↔ ∀ 𝑥 ∈ ( Base ‘ 𝐵 ) ∀ 𝑦 ∈ ( Base ‘ 𝐵 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) )
16 12 15 anbi12d ⊢ ( 𝜑 → ( ( 𝑓 ( 𝐴 Func 𝐶 ) 𝑔 ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐴 ) ∀ 𝑦 ∈ ( Base ‘ 𝐴 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) ↔ ( 𝑓 ( 𝐵 Func 𝐷 ) 𝑔 ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐵 ) ∀ 𝑦 ∈ ( Base ‘ 𝐵 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) ) )
17 eqid ⊢ ( Base ‘ 𝐴 ) = ( Base ‘ 𝐴 )
18 17 isfth ⊢ ( 𝑓 ( 𝐴 Faith 𝐶 ) 𝑔 ↔ ( 𝑓 ( 𝐴 Func 𝐶 ) 𝑔 ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐴 ) ∀ 𝑦 ∈ ( Base ‘ 𝐴 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) )
19 eqid ⊢ ( Base ‘ 𝐵 ) = ( Base ‘ 𝐵 )
20 19 isfth ⊢ ( 𝑓 ( 𝐵 Faith 𝐷 ) 𝑔 ↔ ( 𝑓 ( 𝐵 Func 𝐷 ) 𝑔 ∧ ∀ 𝑥 ∈ ( Base ‘ 𝐵 ) ∀ 𝑦 ∈ ( Base ‘ 𝐵 ) Fun ◡ ( 𝑥 𝑔 𝑦 ) ) )
21 16 18 20 3bitr4g ⊢ ( 𝜑 → ( 𝑓 ( 𝐴 Faith 𝐶 ) 𝑔 ↔ 𝑓 ( 𝐵 Faith 𝐷 ) 𝑔 ) )
22 9 10 21 eqbrrdiv ⊢ ( 𝜑 → ( 𝐴 Faith 𝐶 ) = ( 𝐵 Faith 𝐷 ) )