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Proof step 3 from proof size-two-03 (previous step)

# s0 ≠ 0 ∧ # s0 ≠ 1 → a0 ≠ b ∧ a0 ∈ s0 ∧ b ∈ s0 ⊦ #(s0 ++ a) = 0, #(s0 ++ a) = 1, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 ∈ s0) ∧ (b0 = a ∨ b0 ∈ s0)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ s0, # s0 ≠ 0 ∧ # s0 ≠ 1 → a0 ≠ b ∧ a0 ∈ s0 ∧ b ∈ s0 ⊦ #(s0 ++ a) = 0, #(s0 ++ a) = 1, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 ∈ s0) ∧ (b0 = a ∨ b0 ∈ s0)
Proof step 6:
# s0 ≠ 0 ∧ # s0 ≠ 1 → a0 ≠ b ∧ a0 ∈ s0 ∧ b ∈ s0 ⊦ a ∈ s0, #(s0 ++ a) = 0, #(s0 ++ a) = 1, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 ∈ s0) ∧ (b0 = a ∨ b0 ∈ s0)

Rule argument

fma: a ∈ s0

Additional information

Used formulas: a ∈ s0