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Proof step 17 from proof square-eq (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 2 flags: discard

Additional information

Used simplifier rules: square-abs: ⊦ abs(i * i) = i * i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%square-ls: ⊦ i * i < j * j ↔ abs(i) < abs(j);%irref: ⊦ ¬ i < i; from specification int-basic1