[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 2 from proof mod-gez (previous step)

⊦ 0 ≤ i % j ↔ 0 ≤ i ∨ i % j = 0, j = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: mod-grz-01: ⊦ i < 0 ∧ j ≠ 0 → ¬ 0 < i % j;%mod-zero-06: ⊦ 0 ≤ i ∧ j ≠ 0 → ¬ i % j < 0;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%irref: ⊦ ¬ i < i; from specification int-basic1%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs