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Proof step 11 from proof middle-coef-larger-strongest (previous step)

binom_coef(2 * i, i) ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), 4 ^ i = Σ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), binom_coef(2 * i, i) * 2 * i < 4 ^ i, 0 < i, 0 ≤ i ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 12:
∀ i0. i0 ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)) → i0 ≤ binom_coef(2 * i, i), binom_coef(2 * i, i) ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), 4 ^ i = Σ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), binom_coef(2 * i, i) * 2 * i < 4 ^ i, 0 < i, 0 ≤ i ⊦
Proof step 42:
binom_coef(2 * i, i) ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), 4 ^ i = Σ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)), binom_coef(2 * i, i) * 2 * i < 4 ^ i, 0 < i, 0 ≤ i ⊦ ∀ i0. i0 ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)) → i0 ≤ binom_coef(2 * i, i)

Rule argument

fma: ∀ i0. i0 ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)) → i0 ≤ binom_coef(2 * i, i)

Additional information

Used formulas: ∀ i0. i0 ∈ countFun(0, 2 * i, λ j. binom_coef(2 * i, j)) → i0 ≤ binom_coef(2 * i, i)