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Proof step 7 from proof size-one-02 (previous step)

#(s0 ++ a0) = 1, a ∈ s0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 8:
#(s0 ++ a0) = 1, a ∈ s0, a0 ∈ s0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b
Proof step 10:
#(s0 ++ a0) = 1, a ∈ s0, ∀ a, b. # s0 = 1 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b, a0 ∈ s0

Rule argument

fma: a0 ∈ s0

Additional information

Used formulas: a0 ∈ s0