[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Theorem gcd-divides-both

⊦ j | gcd(i, i0) → j | i ∧ j | i0

Type: A user defined theorem
Validity: The proof is valid.
Features:
Comment:
Used by: gcd-divides-01,
Used by units in proved state:
Proof exists?: Yes

View the proof

Infos about the proof

Proved?: Yes
Interactions: 8
Proof steps: 15
Used lemmas: divides-abs, div-abs-02, div-neg, gcd-same-02, gcd-com, gcd-abs-rew-02, divides-abs-01, gcd-divides-both-h,
Used spec theorems:
- int-abs/Abs-pos
- int-abs/rew-eqzero
Used simplifier rules: