Metamath Proof Explorer


Theorem abelthlem4

Description: Lemma for abelth . (Contributed by Mario Carneiro, 31-Mar-2015)

Ref Expression
Hypotheses abelth.1 φ A : 0
abelth.2 φ seq 0 + A dom
abelth.3 φ M
abelth.4 φ 0 M
abelth.5 S = z | 1 z M 1 z
abelth.6 F = x S n 0 A n x n
Assertion abelthlem4 φ F : S

Proof

Step Hyp Ref Expression
1 abelth.1 φ A : 0
2 abelth.2 φ seq 0 + A dom
3 abelth.3 φ M
4 abelth.4 φ 0 M
5 abelth.5 S = z | 1 z M 1 z
6 abelth.6 F = x S n 0 A n x n
7 nn0uz 0 = 0
8 0zd φ x S 0
9 fveq2 m = n A m = A n
10 oveq2 m = n x m = x n
11 9 10 oveq12d m = n A m x m = A n x n
12 eqid m 0 A m x m = m 0 A m x m
13 ovex A n x n V
14 11 12 13 fvmpt n 0 m 0 A m x m n = A n x n
15 14 adantl φ x S n 0 m 0 A m x m n = A n x n
16 1 adantr φ x S A : 0
17 16 ffvelrnda φ x S n 0 A n
18 5 ssrab3 S
19 18 a1i φ S
20 19 sselda φ x S x
21 expcl x n 0 x n
22 20 21 sylan φ x S n 0 x n
23 17 22 mulcld φ x S n 0 A n x n
24 1 2 3 4 5 abelthlem3 φ x S seq 0 + m 0 A m x m dom
25 7 8 15 23 24 isumcl φ x S n 0 A n x n
26 25 6 fmptd φ F : S