Metamath Proof Explorer


Theorem declt

Description: Comparing two decimal integers (equal higher places). (Contributed by Mario Carneiro, 17-Apr-2015) (Revised by AV, 6-Sep-2021)

Ref Expression
Hypotheses declt.a
|- A e. NN0
declt.b
|- B e. NN0
declt.c
|- C e. NN
declt.l
|- B < C
Assertion declt
|- ; A B < ; A C

Proof

Step Hyp Ref Expression
1 declt.a
 |-  A e. NN0
2 declt.b
 |-  B e. NN0
3 declt.c
 |-  C e. NN
4 declt.l
 |-  B < C
5 10nn
 |-  ; 1 0 e. NN
6 5 1 2 3 4 numlt
 |-  ( ( ; 1 0 x. A ) + B ) < ( ( ; 1 0 x. A ) + C )
7 dfdec10
 |-  ; A B = ( ( ; 1 0 x. A ) + B )
8 dfdec10
 |-  ; A C = ( ( ; 1 0 x. A ) + C )
9 6 7 8 3brtr4i
 |-  ; A B < ; A C