[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

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

∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * Σ countFun(i, i2 + i, intfun) = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ j < i → countFun(i, j, intfun) = []
Proof step 7:
¬ i2 + i < i, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * Σ countFun(i, i2 + i, intfun) = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))
Proof step 15:
i2 + i < i, ∀ j0, i1, i0, intfun. i→n(j0 - i1) < i→n(i2) → true → i0 * Σ countFun(i1, j0, intfun) = Σ countFun(i1, j0, λ j0. i0 * intfun(j0)) ⊦ i0 * Σ [] = Σ countFun(i, i2 + i, λ j0. i0 * intfun(j0))

Rule argument

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

Additional information

Used terms: intfun, i2 + i, i