Metamath Proof Explorer

Theorem redundss3

Description: Implication of redundancy predicate. (Contributed by Peter Mazsa, 26-Oct-2022)

Ref Expression
Hypothesis redundss3.1 
|- D C_ C
Assertion redundss3 
|- ( A Redund <. B , C >. -> A Redund <. B , D >. )

Proof

Step Hyp Ref Expression
1 redundss3.1 
 |-  D C_ C
2 ineq1 
 |-  ( ( A i^i C ) = ( B i^i C ) -> ( ( A i^i C ) i^i D ) = ( ( B i^i C ) i^i D ) )
3 dfss 
 |-  ( D C_ C <-> D = ( D i^i C ) )
4 1 3 mpbi 
 |-  D = ( D i^i C )
5 incom 
 |-  ( D i^i C ) = ( C i^i D )
6 4 5 eqtri 
 |-  D = ( C i^i D )
7 6 ineq2i 
 |-  ( A i^i D ) = ( A i^i ( C i^i D ) )
8 inass 
 |-  ( ( A i^i C ) i^i D ) = ( A i^i ( C i^i D ) )
9 7 8 eqtr4i 
 |-  ( A i^i D ) = ( ( A i^i C ) i^i D )
10 6 ineq2i 
 |-  ( B i^i D ) = ( B i^i ( C i^i D ) )
11 inass 
 |-  ( ( B i^i C ) i^i D ) = ( B i^i ( C i^i D ) )
12 10 11 eqtr4i 
 |-  ( B i^i D ) = ( ( B i^i C ) i^i D )
13 2 9 12 3eqtr4g 
 |-  ( ( A i^i C ) = ( B i^i C ) -> ( A i^i D ) = ( B i^i D ) )
14 13 anim2i 
 |-  ( ( A C_ B /\ ( A i^i C ) = ( B i^i C ) ) -> ( A C_ B /\ ( A i^i D ) = ( B i^i D ) ) )
15 df-redund 
 |-  ( A Redund <. B , C >. <-> ( A C_ B /\ ( A i^i C ) = ( B i^i C ) ) )
16 df-redund 
 |-  ( A Redund <. B , D >. <-> ( A C_ B /\ ( A i^i D ) = ( B i^i D ) ) )
17 14 15 16 3imtr4i 
 |-  ( A Redund <. B , C >. -> A Redund <. B , D >. )