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Proof step 10 from proof countFun-last-01 (previous step)

i1 + i = i2 + 1, 0 ≤ i1, 0 ≤ i, ~1 ≤ i2, i ≤ i2 + 1, ∀ 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, i] flags: rewrite, current

Additional information

Used simplifier rules: countFun-one: ⊦ countFun(i, i, intfun) = intfun(i) ';%Nil: ⊦ [] + ints = ints; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); 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%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%one-03: ⊦ 0 < i ∧ 0 ≤ j → (1 = i + j ↔ i = 1 ∧ j = 0); from specification int-abs%c-04: ⊦ i ' = i0 ' ↔ i = i0; from specification list[intlista] Used terms: intfun, i2, i