Theorem subgrcycl

Description: If a cycle exists in a subgraph of a graph G , then that cycle also exists in G . (Contributed by BTernaryTau, 23-Oct-2023)

	Ref	Expression
Assertion	subgrcycl	$$⊢S\mathrm{SubGraph}G\to \left(F\mathrm{Cycles}\left(S\right)P\to F\mathrm{Cycles}\left(G\right)P\right)$$

Proof

Step	Hyp	Ref	Expression
1		subgrpth	$$⊢S\mathrm{SubGraph}G\to \left(F\mathrm{Paths}\left(S\right)P\to F\mathrm{Paths}\left(G\right)P\right)$$
2	1	anim1d	$$⊢S\mathrm{SubGraph}G\to \left(\left(F\mathrm{Paths}\left(S\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)\to \left(F\mathrm{Paths}\left(G\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)\right)$$
3		iscycl	$$⊢F\mathrm{Cycles}\left(S\right)P\leftrightarrow \left(F\mathrm{Paths}\left(S\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)$$
4		iscycl	$$⊢F\mathrm{Cycles}\left(G\right)P\leftrightarrow \left(F\mathrm{Paths}\left(G\right)P\wedge P\left(0\right)=P\left(\left|F\right|\right)\right)$$
5	2 3 4	3imtr4g	$$⊢S\mathrm{SubGraph}G\to \left(F\mathrm{Cycles}\left(S\right)P\to F\mathrm{Cycles}\left(G\right)P\right)$$