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Proof step 5 from proof countFun-last-01 (previous step)

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) '

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i1 + i, 1, i2] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i1 + i, 1, i2