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

i ≤ i0 + 1, ∀ i1, i2, intfun. i→n(i1 - i2) < i→n(i0 - i) → i2 ≤ i1 + 1 → countFun(i2, i1 + 1, intfun) = countFun(i2, i1, intfun) + intfun(i1 + 1) ' ⊦ countFun(i, i0 + 1, intfun) = countFun(i, i0, intfun) + intfun(i0 + 1) '

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] → [i0, i, i3] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0, i, i3