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Proof step 22 from proof size-three-02 (previous step)

a = c ∨ a = b ∨ a = a0 ⊦ #(∅ ++ a0 ++ b ++ c) = 0, #(∅ ++ a0 ++ b ++ c) = 3, a0 = c, a0 = b, b = c

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: c ∈ ∅ ++ a0 ++ b

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%size-two-01: ⊦ #(∅ ++ a ++ b) = 2 ↔ a ≠ b;% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0; (s) Used formulas: c ∈ ∅ ++ a0 ++ b