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Proof step 1 from proof bint-div2 (conclusion)

⊦ 0 < j0 ∧ j0 ≤ j ∧ j ∈bint → (j0 / 2)∈bint

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j0 ≤ j, ~2147483648 ≤ j, j ≤ 2147483647, 0 < j0, 0 < j, j0 ≤ 2147483647 ⊦ (j0 / 2)∈bint

Rule argument

Additional information

Used simplifier rules: inbint: ⊦ i ∈bint ↔ ~2147483648 ≤ i ∧ i ≤ 2147483647;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs