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Proof step 2 from proof abs-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ abs(ri) = rational(abs(ri.num), ri.denom)
Proof step 4:
0 < j ⊦ rational(abs(rational(i, j).num), rational(i, j).denom) = rational(abs(i), j)

Rule argument

name: abs-def rotate: F subst: [ri] → [rational(i, j)]

Additional information

Used terms: rational(i, j)