Theorem cocan1g 45924

Description: Cancellation law for composition. A version of cocan1 7299 with weaker hypotheses. (Contributed by Eric Schmidt, 29-Sep-2026.)

Hypotheses

Ref	Expression
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

Ref	Expression
cocan1g	⊢ (𝜑 → ((𝐹 ∘ 𝐻) = (𝐹 ∘ 𝐾) ↔ 𝐻 = 𝐾))

Proof of Theorem cocan1g

Step	Hyp	Ref	Expression
1		cocan1g.1	. . . 4 ⊢ (𝜑 → Fun ◡𝐹)
2		cocan1g.2	. . . 4 ⊢ (𝜑 → Rel 𝐻)
3		cocan1g.3	. . . 4 ⊢ (𝜑 → ran 𝐻 ⊆ dom 𝐹)
4	1, 2, 3	cocanss1 45923	. . 3 ⊢ (𝜑 → ((𝐹 ∘ 𝐻) ⊆ (𝐹 ∘ 𝐾) ↔ 𝐻 ⊆ 𝐾))
5		cocan1g.4	. . . 4 ⊢ (𝜑 → Rel 𝐾)
6		cocan1g.5	. . . 4 ⊢ (𝜑 → ran 𝐾 ⊆ dom 𝐹)
7	1, 5, 6	cocanss1 45923	. . 3 ⊢ (𝜑 → ((𝐹 ∘ 𝐾) ⊆ (𝐹 ∘ 𝐻) ↔ 𝐾 ⊆ 𝐻))
8	4, 7	anbi12d 644	. 2 ⊢ (𝜑 → (((𝐹 ∘ 𝐻) ⊆ (𝐹 ∘ 𝐾) ∧ (𝐹 ∘ 𝐾) ⊆ (𝐹 ∘ 𝐻)) ↔ (𝐻 ⊆ 𝐾 ∧ 𝐾 ⊆ 𝐻)))
9		eqss 3946	. 2 ⊢ ((𝐹 ∘ 𝐻) = (𝐹 ∘ 𝐾) ↔ ((𝐹 ∘ 𝐻) ⊆ (𝐹 ∘ 𝐾) ∧ (𝐹 ∘ 𝐾) ⊆ (𝐹 ∘ 𝐻)))
10		eqss 3946	. 2 ⊢ (𝐻 = 𝐾 ↔ (𝐻 ⊆ 𝐾 ∧ 𝐾 ⊆ 𝐻))
11	8, 9, 10	3bitr4g 317	1 ⊢ (𝜑 → ((𝐹 ∘ 𝐻) = (𝐹 ∘ 𝐾) ↔ 𝐻 = 𝐾))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ⊆ wss 3899  ^◡ccnv 5650  dom cdm 5651  ran crn 5652   ∘ ccom 5655  Rel wrel 5656  Fun wfun 6532

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-fun 6540

This theorem is used by: (None)