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Proof step 1 from proof bint-mult2 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs