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Proof step 20 from proof induction (previous step)

f(λ x. μ(f)(x) ∧ φ(x))(x0), μ(f)(x), ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0), 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)

User applied rule all left

And obtained the following new goals

Proof step 21:
f(λ x. μ(f)(x) ∧ φ(x))(x0), μ(f)(x), (λ x. μ(f)(x) ∧ φ(x)) ⊆ μ(f) → f(λ x. μ(f)(x) ∧ φ(x)) ⊆ f(μ(f)), 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)

Rule argument

pos: left 3 terms: λ x. μ(f)(x) ∧ φ(x), μ(f) abort: F discard: T computed: F

Additional information

Used terms: λ x. μ(f)(x) ∧ φ(x), μ(f)