[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,

Proof step 16 from proof mod-by6 (previous step)

((i1 + 1) * (2 + i1) * (2 * (i1 + 1) + 1)) % 6 = 0, 0 ≤ i1 ⊦ (i1 * (i1 + 1) * (2 * i1 + 1)) % 6 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: modby6-help seq: ⊦ (i + 1) * (2 + i) * (2 * (i + 1) + 1) = 6 * (i + 1) * (i + 1) + i * (i + 1) * (2 * i + 1) subst: [i] → [i1] flags: rewrite, current

Additional information

Used terms: i1