Metamath Proof Explorer

Theorem ex-natded5.2

Description: Theorem 5.2 of Clemente p. 15, translated line by line using the interpretation of natural deduction in Metamath. For information about ND and Metamath, see the page on Deduction Form and Natural Deduction in Metamath Proof Explorer. The original proof, which uses Fitch style, was written as follows:

#	MPE#	ND Expression	MPE Translation	ND Rationale	MPE Rationale
1	5	( ( ps /\ ch ) -> th )	( ph -> ( ( ps /\ ch ) -> th ) )	Given	$e.
2	2	( ch -> ps )	( ph -> ( ch -> ps ) )	Given	$e.
3	1	ch	( ph -> ch )	Given	$e.
4	3	ps	( ph -> ps )	->E 2,3	mpd , the MPE equivalent of ->E, 1,2
5	4	( ps /\ ch )	( ph -> ( ps /\ ch ) )	/\I 4,3	jca , the MPE equivalent of /\I, 3,1
6	6	th	( ph -> th )	->E 1,5	mpd , the MPE equivalent of ->E, 4,5

The original used Latin letters for predicates; we have replaced them with Greek letters to follow Metamath naming conventions and so that it is easier to follow the Metamath translation. The Metamath line-for-line translation of this natural deduction approach precedes every line with an antecedent including ph and uses the Metamath equivalents of the natural deduction rules. Below is the final Metamath proof (which reorders some steps). A much more efficient proof, using more of Metamath and MPE's capabilities, is shown in ex-natded5.2-2 . A proof without context is shown in ex-natded5.2i . (Contributed by Mario Carneiro, 9-Feb-2017) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Hypotheses	ex-natded5.2.1	\|- ( ph -> ( ( ps /\ ch ) -> th ) )
	ex-natded5.2.2	\|- ( ph -> ( ch -> ps ) )
	ex-natded5.2.3	\|- ( ph -> ch )
Assertion	ex-natded5.2	\|- ( ph -> th )

Proof

Step	Hyp	Ref	Expression
1		ex-natded5.2.1	\|- ( ph -> ( ( ps /\ ch ) -> th ) )
2		ex-natded5.2.2	\|- ( ph -> ( ch -> ps ) )
3		ex-natded5.2.3	\|- ( ph -> ch )
4	3 2	mpd	\|- ( ph -> ps )
5	4 3	jca	\|- ( ph -> ( ps /\ ch ) )
6	5 1	mpd	\|- ( ph -> th )