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Proof step 6 from proof countFun-last-01 (previous step)

i1 + i = i2 + 1, 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, i2, intfun) + intfun(i1 + i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
i1 + i = i2 + 1, 0 ≤ i1, 0 ≤ i, ~1 ≤ i2, ∀ 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, i2 + 1, intfun) = countFun(i, i2, intfun) + intfun(i2 + 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%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs