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Proof step 11 from proof size-three (previous step)

2 < #(∅ ++ a ++ b), # s ≤ #(∅ ++ a ++ b), a ∈ s, 2 < # s, b ∈ s ⊦ b = a, # s = 0, #(∅ ++ a ++ b) = 0, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: b ∈ ∅ ++ a

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%size-one-01: ⊦ #(∅ ++ a) = 1;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%size-one-01: ⊦ #(∅ ++ a) = 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used formulas: b ∈ ∅ ++ a