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Theorem ex-natded5.3 20768
Description: Theorem 5.3 of [Laboreo] p. 16, translated line by line using an interpretation of natural deduction in Metamath. A much more efficient proof, using more of Metamath and MPE's capabilities, is shown in ex-natded5.3-2 20769. A proof without context is shown in ex-natded5.3i 20770. 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 TranslationND Rationale MPE Rationale
12;3  ( ps  ->  ch )  ( ph  ->  ( ps  ->  ch ) ) Given $e; adantr 453 to move it into the ND hypothesis
25;6  ( ch  ->  th )  ( ph  ->  ( ch  ->  th ) ) Given $e; adantr 453 to move it into the ND hypothesis
31 ...|  ps  ( ( ph  /\  ps )  ->  ps ) ND hypothesis assumption simpr 449, to access the new assumption
44 ...  ch  ( ( ph  /\  ps )  ->  ch )  ->E 1,3 mpd 16, the MPE equivalent of  ->E, 1.3. adantr 453 was used to transform its dependency (we could also use imp 420 to get this directly from 1)
57 ...  th  ( ( ph  /\  ps )  ->  th )  ->E 2,4 mpd 16, the MPE equivalent of  ->E, 4,6. adantr 453 was used to transform its dependency
68 ...  ( ch  /\  th )  ( ( ph  /\  ps )  ->  ( ch  /\  th ) )  /\I 4,5 jca 520, the MPE equivalent of  /\I, 4,7
79  ( ps  ->  ( ch  /\  th ) )  ( ph  ->  ( ps  ->  ( ch  /\  th ) ) )  ->I 3,6 ex 425, the MPE equivalent of  ->I, 8

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. (Proof modification is discouraged.) (Contributed by Mario Carneiro, 9-Feb-2017.)

Hypotheses
Ref Expression
ex-natded5.3.1  |-  ( ph  ->  ( ps  ->  ch ) )
ex-natded5.3.2  |-  ( ph  ->  ( ch  ->  th )
)
Assertion
Ref Expression
ex-natded5.3  |-  ( ph  ->  ( ps  ->  ( ch  /\  th ) ) )

Proof of Theorem ex-natded5.3
StepHypRef Expression
1 simpr 449 . . . 4  |-  ( (
ph  /\  ps )  ->  ps )
2 ex-natded5.3.1 . . . . 5  |-  ( ph  ->  ( ps  ->  ch ) )
32adantr 453 . . . 4  |-  ( (
ph  /\  ps )  ->  ( ps  ->  ch ) )
41, 3mpd 16 . . 3  |-  ( (
ph  /\  ps )  ->  ch )
5 ex-natded5.3.2 . . . . 5  |-  ( ph  ->  ( ch  ->  th )
)
65adantr 453 . . . 4  |-  ( (
ph  /\  ps )  ->  ( ch  ->  th )
)
74, 6mpd 16 . . 3  |-  ( (
ph  /\  ps )  ->  th )
84, 7jca 520 . 2  |-  ( (
ph  /\  ps )  ->  ( ch  /\  th ) )
98ex 425 1  |-  ( ph  ->  ( ps  ->  ( ch  /\  th ) ) )
Colors of variables: wff set class
Syntax hints:    -> wi 6    /\ wa 360
This theorem was proved from axioms:  ax-1 7  ax-2 8  ax-3 9  ax-mp 10
This theorem depends on definitions:  df-bi 179  df-an 362
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