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Proof step 15 from proof count-rec (previous step)

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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0, j0] → [i, i2]

Additional information

Used simplifier rules: labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%abssub: ⊦ j ≤ i → abs(i - j) = i - j; from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%lep-01: ⊦ i + j ≤ i ↔ j ≤ 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 Used terms: i, i2