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

⊦ abs(rational(i, i0)).num = abs(rational(i, i0).num)

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:
⊦ rational(abs(rational(i, i0).num), rational(i, i0).denom).num = abs(rational(i, i0).num)

Rule argument

name: abs-def rotate: F subst: [ri] → [rational(i, i0)]

Additional information

Used terms: rational(i, i0)