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Proof step 2 from proof bint-div2-01 (previous step)

j0 ≤ j, j ∈bint, 0 < j0, 0 < j ⊦ bi2i(i2bi(j0 / 2)) = j0 / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ∈bint → bi2i(i2bi(i)) = i

Rule argument

spec: inst: name: biibi-in seq: ⊦ i ∈bint → bi2i(i2bi(i)) = i subst: [i] → [j0 / 2] flags: rewrite, current

Additional information

Used simplifier rules: bint-div2: ⊦ 0 < j0 ∧ j0 ≤ j ∧ j ∈bint → (j0 / 2)∈bint; Used terms: j0 / 2