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Proof step 2 from proof countFun-last (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
j = i0 + 1, i ≤ j ⊦ countFun(i, j, intfun) = countFun(i, i0, intfun) + intfun(j) '

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i0] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i0