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Proof step 26 from proof induction (
previous step
)
∀ x. f(λ x. μ(f)(x) ∧ φ(x))(x) → f(μ(f))(x), f(λ x. μ(f)(x) ∧ φ(x))(x0), μ(f)(x), φ(x0), ∀ x. f(μ(f))(x) → μ(f)(x), ∀ x. f(λ x. μ(f)(x) ∧ φ(x))(x) → φ(x) ⊦ φ(x), μ(f)(x0), f(μ(f))(x0)
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: x0
Additional information
Used terms: x0