Theorem decaddc

Description: Add two numerals M and N (with carry). (Contributed by Mario Carneiro, 18-Feb-2014) (Revised by AV, 6-Sep-2021)

	Ref	Expression
Hypotheses	decma.a	|- A e. NN0
	decma.b	|- B e. NN0
	decma.c	|- C e. NN0
	decma.d	|- D e. NN0
	decma.m	|- M = ; A B
	decma.n	|- N = ; C D
	decaddc.e	|- ( ( A + C ) + 1 ) = E
	decaddc.f	|- F e. NN0
	decaddc.2	|- ( B + D ) = ; 1 F
Assertion	decaddc	|- ( M + N ) = ; E F

Proof

Step	Hyp	Ref	Expression
1		decma.a	|- A e. NN0
2		decma.b	|- B e. NN0
3		decma.c	|- C e. NN0
4		decma.d	|- D e. NN0
5		decma.m	|- M = ; A B
6		decma.n	|- N = ; C D
7		decaddc.e	|- ( ( A + C ) + 1 ) = E
8		decaddc.f	|- F e. NN0
9		decaddc.2	|- ( B + D ) = ; 1 F
10		10nn0	|- ; 1 0 e. NN0
11		dfdec10	|- ; A B = ( ( ; 1 0 x. A ) + B )
12	5 11	eqtri	|- M = ( ( ; 1 0 x. A ) + B )
13		dfdec10	|- ; C D = ( ( ; 1 0 x. C ) + D )
14	6 13	eqtri	|- N = ( ( ; 1 0 x. C ) + D )
15		dfdec10	|- ; 1 F = ( ( ; 1 0 x. 1 ) + F )
16	9 15	eqtri	|- ( B + D ) = ( ( ; 1 0 x. 1 ) + F )
17	10 1 2 3 4 12 14 8 7 16	numaddc	|- ( M + N ) = ( ( ; 1 0 x. E ) + F )
18		dfdec10	|- ; E F = ( ( ; 1 0 x. E ) + F )
19	17 18	eqtr4i	|- ( M + N ) = ; E F