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Proof step 4 from proof countFun-last-01 (previous step)

j = i1 + i, i ≤ j, 0 ≤ i, 0 ≤ j, ∀ j0, i0, intfun. i→n(j0 - i0) < i→n(i1) → true → i0 ≤ j0 ∧ 0 ≤ i0 → countFun(i0, j0, intfun) = countFun(i0, j0 - 1, intfun) + intfun(j0) ' ⊦ countFun(i, j, intfun) = countFun(i, j - 1, intfun) + intfun(j) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 ≤ i, 0 ≤ i1 + i, 0 ≤ i1, ∀ j0, i0, intfun. i→n(j0 - i0) < i→n(i1) → true → i0 ≤ j0 ∧ 0 ≤ i0 → countFun(i0, j0, intfun) = countFun(i0, j0 - 1, intfun) + intfun(j0) ' ⊦ countFun(i, i1 + i, intfun) = countFun(i, i1 + i - 1, intfun) + intfun(i1 + i) '

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs