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Proof step 10 from proof subsize-eq (previous step)

# s = # s0, ∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# s = # s1 + 1, ∀ a0. a0 ∈ s → a0 ∈ s1 ⊦ a ∈ s, a ∈ s1

Rule argument

Additional information

Used simplifier rules: Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);