Metamath Proof Explorer


Theorem dif1card

Description: The cardinality of a nonempty finite set is one greater than the cardinality of the set with one element removed. (Contributed by Jeff Madsen, 2-Sep-2009) (Proof shortened by Mario Carneiro, 2-Feb-2013)

Ref Expression
Assertion dif1card A Fin X A card A = suc card A X

Proof

Step Hyp Ref Expression
1 diffi A Fin A X Fin
2 isfi A X Fin m ω A X m
3 simp3 X A m ω A X m A X m
4 en2sn X A m ω X m
5 4 3adant3 X A m ω A X m X m
6 disjdifr A X X =
7 6 a1i X A m ω A X m A X X =
8 nnord m ω Ord m
9 ordirr Ord m ¬ m m
10 8 9 syl m ω ¬ m m
11 disjsn m m = ¬ m m
12 10 11 sylibr m ω m m =
13 12 3ad2ant2 X A m ω A X m m m =
14 unen A X m X m A X X = m m = A X X m m
15 3 5 7 13 14 syl22anc X A m ω A X m A X X m m
16 difsnid X A A X X = A
17 df-suc suc m = m m
18 17 eqcomi m m = suc m
19 18 a1i X A m m = suc m
20 16 19 breq12d X A A X X m m A suc m
21 20 3ad2ant1 X A m ω A X m A X X m m A suc m
22 15 21 mpbid X A m ω A X m A suc m
23 peano2 m ω suc m ω
24 23 3ad2ant2 X A m ω A X m suc m ω
25 cardennn A suc m suc m ω card A = suc m
26 22 24 25 syl2anc X A m ω A X m card A = suc m
27 cardennn A X m m ω card A X = m
28 27 ancoms m ω A X m card A X = m
29 28 3adant1 X A m ω A X m card A X = m
30 suceq card A X = m suc card A X = suc m
31 29 30 syl X A m ω A X m suc card A X = suc m
32 26 31 eqtr4d X A m ω A X m card A = suc card A X
33 32 3expib X A m ω A X m card A = suc card A X
34 33 com12 m ω A X m X A card A = suc card A X
35 34 rexlimiva m ω A X m X A card A = suc card A X
36 2 35 sylbi A X Fin X A card A = suc card A X
37 1 36 syl A Fin X A card A = suc card A X
38 37 imp A Fin X A card A = suc card A X