[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 21 from proof square-eq (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 22:
i + abs(j) = 0, i = ~ abs(j), i < 0 ⊦ i * i = j * j, j = 0

Rule argument

fma: i = ~ abs(j)

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i = ~ abs(j)