Metamath Proof Explorer


Theorem cocan1g

Description: Cancellation law for composition. A version of cocan1 with weaker hypotheses. (Contributed by Eric Schmidt, 29-Sep-2026)

Ref Expression
Hypotheses cocan1g.1 ⊢ ( 𝜑 → Fun ◡ 𝐹 )
cocan1g.2 ⊢ ( 𝜑 → Rel 𝐻 )
cocan1g.3 ⊢ ( 𝜑 → ran 𝐻 ⊆ dom 𝐹 )
cocan1g.4 ⊢ ( 𝜑 → Rel 𝐾 )
cocan1g.5 ⊢ ( 𝜑 → ran 𝐾 ⊆ dom 𝐹 )
Assertion cocan1g ( 𝜑 → ( ( 𝐹 ∘ 𝐻 ) = ( 𝐹 ∘ 𝐾 ) ↔ 𝐻 = 𝐾 ) )

Proof

Step Hyp Ref Expression
1 cocan1g.1 ⊢ ( 𝜑 → Fun ◡ 𝐹 )
2 cocan1g.2 ⊢ ( 𝜑 → Rel 𝐻 )
3 cocan1g.3 ⊢ ( 𝜑 → ran 𝐻 ⊆ dom 𝐹 )
4 cocan1g.4 ⊢ ( 𝜑 → Rel 𝐾 )
5 cocan1g.5 ⊢ ( 𝜑 → ran 𝐾 ⊆ dom 𝐹 )
6 1 2 3 cocanss1 ⊢ ( 𝜑 → ( ( 𝐹 ∘ 𝐻 ) ⊆ ( 𝐹 ∘ 𝐾 ) ↔ 𝐻 ⊆ 𝐾 ) )
7 1 4 5 cocanss1 ⊢ ( 𝜑 → ( ( 𝐹 ∘ 𝐾 ) ⊆ ( 𝐹 ∘ 𝐻 ) ↔ 𝐾 ⊆ 𝐻 ) )
8 6 7 anbi12d ⊢ ( 𝜑 → ( ( ( 𝐹 ∘ 𝐻 ) ⊆ ( 𝐹 ∘ 𝐾 ) ∧ ( 𝐹 ∘ 𝐾 ) ⊆ ( 𝐹 ∘ 𝐻 ) ) ↔ ( 𝐻 ⊆ 𝐾 ∧ 𝐾 ⊆ 𝐻 ) ) )
9 eqss ⊢ ( ( 𝐹 ∘ 𝐻 ) = ( 𝐹 ∘ 𝐾 ) ↔ ( ( 𝐹 ∘ 𝐻 ) ⊆ ( 𝐹 ∘ 𝐾 ) ∧ ( 𝐹 ∘ 𝐾 ) ⊆ ( 𝐹 ∘ 𝐻 ) ) )
10 eqss ⊢ ( 𝐻 = 𝐾 ↔ ( 𝐻 ⊆ 𝐾 ∧ 𝐾 ⊆ 𝐻 ) )
11 8 9 10 3bitr4g ⊢ ( 𝜑 → ( ( 𝐹 ∘ 𝐻 ) = ( 𝐹 ∘ 𝐾 ) ↔ 𝐻 = 𝐾 ) )