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Proof step 19 from proof count-i2j-ordered (previous step)

count-i2j(i + 1, i1 + i) = i2 ' + ints0, i2 ≤ i1 + i, i < i2, i ≤ i2, 0 < i1, ∀ j0, i0. i→n(j0 - i0) < i→n(i1) → true → ordered≤(count-i2j(i0, j0)) ⊦ i1 = 0, ordered≤(i2 ' + ints0)

User applied rule apply induction

And closed the goal!

Rule argument

subst: [j0, i0] → [i1 + i, i + 1]

Additional information

Used simplifier rules: sub-add-05: ⊦ i + j - i + 1 = j - 1; 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%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: i1 + i, i + 1