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Proof step 3 from proof cast-law-compl (previous step)

0 < i ⊦ cast(i, ~ j - 1) = cast(i, bcompl(cast(i, j)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ bcompl(i) = ~ i - 1
Proof step 5:
0 < i ⊦ cast(i, ~ j - 1) = cast(i, ~ cast(i, j) - 1)

Rule argument

name: compl-def rotate: F subst: [i] → [cast(i, j)]

Additional information

Used terms: cast(i, j)