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Proof step 1 from proof countFun-at (conclusion)

⊦ n→i(n) ≤ j - i ∧ 0 ≤ i ∧ i ≤ j → countFun(i, j, intfun)[n] = intfun(i + n→i(n))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
i ≤ j, n→i(n) + i ≤ j, 0 ≤ i, 0 ≤ j ⊦ countFun(i, j, intfun)[n] = intfun(i + n→i(n))

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; from specification int-abs