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Proof step 16 from proof countFun-last (previous step)

0 ≤ i3, ∀ i1, i2, intfun. i→n(i1 - i2) < i→n(i3) → i2 ≤ i1 + 1 → countFun(i2, i1 + 1, intfun) = countFun(i2, i1, intfun) + intfun(i1 + 1) ' ⊦ countFun(i + 1, (i3 + i) + 1, intfun) = countFun(i + 1, i3 + i, intfun) + intfun((i3 + i) + 1) '

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i1, i2, intfun] → [i + i3, i + 1, intfun]

Additional information

Used simplifier rules: countFun-one: ⊦ countFun(i, i, intfun) = intfun(i) ';%countFun-base-01: ⊦ countFun(i + 1, i, intfun) = [];%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%le-add: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%sub-add-05: ⊦ i + j - i + 1 = j - 1; from specification int-abs%abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%Nil: ⊦ [] + ints = ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%c-04: ⊦ i ' = i0 ' ↔ i = i0; from specification list[intlista] Used terms: i + i3, i + 1, intfun