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Proof step 13 from proof subsize-eq (previous step)

∃ a. a ∈ s ∧ ¬ a ∈ s1, # s = # s1 + 1, ∀ a0. a0 ∈ s → a0 ∈ s1 ⊦ a ∈ s, a ∈ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# s = # s1 + 1, a0 ∈ s, ∀ a0. a0 ∈ s → a0 ∈ s1 ⊦ a0 ∈ s1, a ∈ s1, a ∈ s

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);