Metamath Proof Explorer


Theorem plydivlem1

Description: Lemma for plydivalg . (Contributed by Mario Carneiro, 24-Jul-2014)

Ref Expression
Hypotheses plydiv.pl φ x S y S x + y S
plydiv.tm φ x S y S x y S
plydiv.rc φ x S x 0 1 x S
plydiv.m1 φ 1 S
Assertion plydivlem1 φ 0 S

Proof

Step Hyp Ref Expression
1 plydiv.pl φ x S y S x + y S
2 plydiv.tm φ x S y S x y S
3 plydiv.rc φ x S x 0 1 x S
4 plydiv.m1 φ 1 S
5 1pneg1e0 1 + -1 = 0
6 neg1mulneg1e1 -1 -1 = 1
7 2 4 4 caovcld φ -1 -1 S
8 6 7 eqeltrrid φ 1 S
9 1 8 4 caovcld φ 1 + -1 S
10 5 9 eqeltrrid φ 0 S