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Proof step 3 from proof Binom-theorem-countFun (previous step)

0 ≤ j, ∀ j0, i, i0. i→n(j0) < i→n(j) → true → 0 ≤ j0 → (i + i0) ^ j0 = Σ countFun(0, j0, λ j1. binom_coef(j0, j1) * i ^ j1 * i0 ^ (j0 - j1)) ⊦ (i + i0) ^ j = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

User applied rule cut formula

And obtained the following new goals

Proof step 4:
0 ≤ j, ∀ j0, i, i0. i→n(j0) < i→n(j) → true → 0 ≤ j0 → (i + i0) ^ j0 = Σ countFun(0, j0, λ j1. binom_coef(j0, j1) * i ^ j1 * i0 ^ (j0 - j1)) ⊦ j = 0, (i + i0) ^ j = Σ countFun(0, j, λ j0. binom_coef(j, j0) * i ^ j0 * i0 ^ (j - j0))

Rule argument

fma: j = 0

Additional information

Used simplifier rules: coef-zero: ⊦ 0 ≤ i → binom_coef(i, 0) = 1;%Pot-zero: ⊦ j ^ 0 = 1; from specification int-pot%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%countFun-one: ⊦ countFun(i, i, intfun) = intfun(i) '; from specification intlist%commutativity: ⊦ i * j = j * i; from specification int-mult%sum-one: ⊦ Σ i ' = i; from specification intlist Used formulas: j = 0