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Proof step 6 from proof cast-law-compl (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ~(i + 1) = ~ i - 1
Proof step 8:
0 < i ⊦ cast(i, ~ j - 1) = cast(i, ~(cast(i, j) + 1))

Rule argument

seq: ⊦ ~(i + 1) = ~ i - 1 subst: [i] → [j] rotate: F spec: int-abs inst: name: minus-01

Additional information

Used terms: j