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Proof step 2 from proof bint-mult2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i ∈bint → bi2i(i2bi(i)) = i
Proof step 4:
¬ j0 ∈bint, j0 * 2 ≤ j, j ∈bint, 0 < j0 ⊦ bi2i(i2bi(j0)) = j0

Rule argument

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

Additional information

Used terms: j0