Theorem cocan1g 46061

Description: Cancellation law for composition. A version of cocan1 7303 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 46060	. . 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 46060	. . 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 3924	. 2 ⊢ ((𝐹 ∘ 𝐻) = (𝐹 ∘ 𝐾) ↔ ((𝐹 ∘ 𝐻) ⊆ (𝐹 ∘ 𝐾) ∧ (𝐹 ∘ 𝐾) ⊆ (𝐹 ∘ 𝐻)))
10		eqss 3924	. 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 1569   ⊆ wss 3896  ^◡ccnv 5650  dom cdm 5651  ran crn 5652   ∘ ccom 5655  Rel wrel 5656  Fun wfun 6537

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  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 6545

This theorem is used by: (None)