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Proof step 1 from proof count-empty (conclusion)

⊦ count-i2j(j, j - 1) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ j < i → count-i2j(i, j) = []

Rule argument

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

Additional information

Used simplifier rules: count-i2j-empty-02: ⊦ count-i2j(i, j) = [] ↔ j < i;%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j - 1, j