Theorem simprld

Description: Deduction eliminating a double conjunct. (Contributed by Glauco Siliprandi, 11-Dec-2019)

	Ref	Expression
Hypothesis	simprld.1	|- ( ph -> ( ps /\ ( ch /\ th ) ) )
Assertion	simprld	|- ( ph -> ch )

Proof

Step	Hyp	Ref	Expression
1		simprld.1	|- ( ph -> ( ps /\ ( ch /\ th ) ) )
2	1	simprd	|- ( ph -> ( ch /\ th ) )
3	2	simpld	|- ( ph -> ch )