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Proof step 6 from proof firstn-restn-in (previous step)

n0 ≤ # y, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ y ↔ a ∈ firstn(n0, y) ∨ a ∈ restn(n0, y), a = a0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a ∈ y, n0 ≤ # y, ∀ x, a. n0 ≤ # x → (a ∈ x ↔ a ∈ firstn(n0, x) ∨ a ∈ restn(n0, x)) ⊦ a ∈ y ↔ a ∈ firstn(n0, y) ∨ a ∈ restn(n0, y), a = a0

Rule argument

fma: a ∈ y

Additional information

Used simplifier rules: firstN-in: ⊦ ¬ a ∈ x ∧ n ≤ # x → ¬ a ∈ firstn(n, x);%rest-in-01: ⊦ a ∈ restn(n, x) ∧ ¬ a ∈ x → ¬ n ≤ # x;%rest-in: ⊦ a ∈ restn(n, x) ∧ n ≤ # x → a ∈ x; Used formulas: a ∈ y