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Proof step 7 from proof countFun-last-01 (previous step)

i1 + i = i2 + 1, 0 ≤ i1, 0 ≤ i, ~1 ≤ i2, ∀ 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, i2 + 1, intfun) = countFun(i, i2, intfun) + intfun(i2 + 1) '

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: e-02: ⊦ ints + i ' ≠ []; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%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 + 1, i