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Proof step 5 from proof bint-mult2 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: bint-trans-pos seq: ⊦ j ∈bint ∧ 0 < i ∧ i ≤ j → (¬ i ∈bint ↔ false) subst: [j, i] → [j, j0] flags: rewrite, current

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: j, j0