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Proof step 4 from proof size-exgr (previous step)

s0 = s ++ a0 ∧ s0 ≠ s ⊦ s0 = ∅, # s0 = 0, ∃ a. a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ a0 ∈ s, ∃ a. a = a0 ∨ a ∈ s

Rule argument

Additional information

Used simplifier rules: ne: ⊦ ∅ ≠ s ++ a;%eq-03: ⊦ s ++ a = s ↔ a ∈ s;%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2