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Proof step 3 from proof rev-in (previous step)

∀ a. a ∈ rev(x0) ↔ a ∈ x0 ⊦ a ∈ rev(x0) ↔ a ∈ x0, a = a0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ x0, ∀ a. a ∈ rev(x0) ↔ a ∈ x0 ⊦ a ∈ rev(x0) ↔ a ∈ x0, a = a0
Proof step 6:
∀ a. a ∈ rev(x0) ↔ a ∈ x0 ⊦ a ∈ x0, a ∈ rev(x0) ↔ a ∈ x0, a = a0

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0