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Proof step 4 from proof del (previous step)

∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1), a = a0 ∨ a ∈ s0 ⊦ ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a0 ∈ s0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1), a = a0 ∨ a ∈ s0 ⊦ ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1
Proof step 7:
∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1), a = a0 ∨ a ∈ s0 ⊦ a0 ∈ s0, ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

fma: a0 ∈ s0

Additional information

Used formulas: a0 ∈ s0