[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 4 from proof absa (previous step)

i + j < 0 ⊦ abs(i + j) = abs(i) + abs(j), 0 < i, 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
i + j < 0 ⊦ 0 = (abs(i) + abs(j)) + i + j, 0 < j, 0 < i

Rule argument

Additional information

Used simplifier rules: Abs-neg: ⊦ j < 0 → abs(j) = 0 - j;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;