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Proof step 14 from proof count-i2j-ordered (previous step)

∀ j0, i0. i→n(j0 - i0) < i→n(0) → true → ordered≤(count-i2j(i0, j0)) ⊦ ordered≤(i ' + count-i2j(i + 1, i))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: counti2j-empty seq: ⊦ j < i → count-i2j(i, j) = [] subst: [j, i] → [i, i + 1] flags: rewrite, current

Additional information

Used simplifier rules: succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%ler: ⊦ i ≤ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%app: ⊦ ints + [] = ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%le-o: ⊦ ordered≤(i '); from specification olist[intlista]%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista] Used terms: i, i + 1