Metamath Proof Explorer


Theorem cocanss1

Description: Cancellation law for composition. Suggested by BJ. (Contributed by Eric Schmidt, 30-Sep-2026)

Ref Expression
Hypotheses cocan1g.1 ⊢ φ → Fun ⁡ F -1
cocan1g.2 ⊢ φ → Rel ⁡ H
cocan1g.3 ⊢ φ → ran ⁡ H ⊆ dom ⁡ F
Assertion cocanss1 ⊢ φ → F ∘ H ⊆ F ∘ K ↔ H ⊆ K

Proof

Step Hyp Ref Expression
1 cocan1g.1 ⊢ φ → Fun ⁡ F -1
2 cocan1g.2 ⊢ φ → Rel ⁡ H
3 cocan1g.3 ⊢ φ → ran ⁡ H ⊆ dom ⁡ F
4 relcnv ⊢ Rel ⁡ H -1
5 4 a1i ⊢ φ → Rel ⁡ H -1
6 df-rn ⊢ ran ⁡ H = dom ⁡ H -1
7 dfdm4 ⊢ dom ⁡ F = ran ⁡ F -1
8 3 6 7 3sstr3g ⊢ φ → dom ⁡ H -1 ⊆ ran ⁡ F -1
9 1 5 8 cocanss2 ⊢ φ → H -1 ∘ F -1 ⊆ K -1 ∘ F -1 ↔ H -1 ⊆ K -1
10 relco ⊢ Rel ⁡ F ∘ H
11 cnvssb ⊢ Rel ⁡ F ∘ H → F ∘ H ⊆ F ∘ K ↔ F ∘ H -1 ⊆ F ∘ K -1
12 10 11 ax-mp ⊢ F ∘ H ⊆ F ∘ K ↔ F ∘ H -1 ⊆ F ∘ K -1
13 cnvco ⊢ F ∘ H -1 = H -1 ∘ F -1
14 cnvco ⊢ F ∘ K -1 = K -1 ∘ F -1
15 13 14 sseq12i ⊢ F ∘ H -1 ⊆ F ∘ K -1 ↔ H -1 ∘ F -1 ⊆ K -1 ∘ F -1
16 12 15 bitri ⊢ F ∘ H ⊆ F ∘ K ↔ H -1 ∘ F -1 ⊆ K -1 ∘ F -1
17 16 a1i ⊢ φ → F ∘ H ⊆ F ∘ K ↔ H -1 ∘ F -1 ⊆ K -1 ∘ F -1
18 cnvssb ⊢ Rel ⁡ H → H ⊆ K ↔ H -1 ⊆ K -1
19 2 18 syl ⊢ φ → H ⊆ K ↔ H -1 ⊆ K -1
20 9 17 19 3bitr4d ⊢ φ → F ∘ H ⊆ F ∘ K ↔ H ⊆ K