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Proof step 6 from proof size-exgr (previous step)

#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1 ++ a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#(s1 ++ a) < # s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ ∃ a0. a0 ∈ s0 ∧ a0 ≠ a ∧ ¬ a0 ∈ s1

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic