Metamath Proof Explorer


Theorem decltc

Description: Comparing two decimal integers (unequal higher places). (Contributed by Mario Carneiro, 18-Feb-2014) (Revised by AV, 6-Sep-2021)

Ref Expression
Hypotheses declt.a
|- A e. NN0
declt.b
|- B e. NN0
decltc.c
|- C e. NN0
decltc.d
|- D e. NN0
decltc.s
|- C < ; 1 0
decltc.l
|- A < B
Assertion decltc
|- ; A C < ; B D

Proof

Step Hyp Ref Expression
1 declt.a
 |-  A e. NN0
2 declt.b
 |-  B e. NN0
3 decltc.c
 |-  C e. NN0
4 decltc.d
 |-  D e. NN0
5 decltc.s
 |-  C < ; 1 0
6 decltc.l
 |-  A < B
7 10nn
 |-  ; 1 0 e. NN
8 7 1 2 3 4 5 6 numltc
 |-  ( ( ; 1 0 x. A ) + C ) < ( ( ; 1 0 x. B ) + D )
9 dfdec10
 |-  ; A C = ( ( ; 1 0 x. A ) + C )
10 dfdec10
 |-  ; B D = ( ( ; 1 0 x. B ) + D )
11 8 9 10 3brtr4i
 |-  ; A C < ; B D