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Proof step 8 from proof countFun-in (previous step)

0 ≤ i2, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → (i0 ∈ countFun(i1, j0, intfun) ↔ (∃ j1. i1 ≤ j1 ∧ j1 ≤ j0 ∧ intfun(j1) = i0)) ⊦ i0 ∈ countFun(i, i2 + i, intfun) ↔ (∃ j0. i ≤ j0 ∧ j0 ≤ i2 + i ∧ intfun(j0) = i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: intlist inst: name: countFun-rec seq: ⊦ i ≤ j → countFun(i, j, intfun) = intfun(i) ' + countFun(i + 1, j, intfun) subst: [intfun, j, i] → [intfun, i2 + i, i] flags: rewrite, current

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: intfun, i2 + i, i