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Proof step 2 from proof size-less (previous step)

# s + 1 < #(s ++ a) ⊦ #(s ++ a) = 0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2 Used formulas: a ∈ s