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Proof step 15 from proof countFun-last (previous step)

i ≤ (i3 + i) + 1, 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) ' ⊦ intfun(i) ' + countFun(i + 1, (i3 + i) + 1, intfun) = (intfun(i) ' + countFun(i + 1, i3 + i, intfun)) + intfun((i3 + i) + 1) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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) '

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%app-01: ⊦ ints + ints0 = ints + ints1 ↔ ints0 = ints1; from specification list[intlista]