Theorem pthisspthorcycl

Description: A path is either a simple path or a cycle (or both). (Contributed by BTernaryTau, 20-Oct-2023)

	Ref	Expression
Assertion	pthisspthorcycl	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(F\mathrm{SPaths}\left(G\right)P\vee F\mathrm{Cycles}\left(G\right)P\right)$$

Proof

Step	Hyp	Ref	Expression
1		pthdepisspth	$$⊢\left(F\mathrm{Paths}\left(G\right)P\wedge P\left(0\right)\ne P\left(\left|F\right|\right)\right)\to F\mathrm{SPaths}\left(G\right)P$$
2	1	ex	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(P\left(0\right)\ne P\left(\left|F\right|\right)\to F\mathrm{SPaths}\left(G\right)P\right)$$
3	2	necon1bd	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(\neg F\mathrm{SPaths}\left(G\right)P\to P\left(0\right)=P\left(\left|F\right|\right)\right)$$
4	3	anc2li	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(\neg F\mathrm{SPaths}\left(G\right)P\to \left(F\mathrm{Paths}\left(G\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)\right)$$
5		iscycl	$$⊢F\mathrm{Cycles}\left(G\right)P\leftrightarrow \left(F\mathrm{Paths}\left(G\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)$$
6	4 5	imbitrrdi	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(\neg F\mathrm{SPaths}\left(G\right)P\to F\mathrm{Cycles}\left(G\right)P\right)$$
7	6	orrd	$$⊢F\mathrm{Paths}\left(G\right)P\to \left(F\mathrm{SPaths}\left(G\right)P\vee F\mathrm{Cycles}\left(G\right)P\right)$$