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Proof step 9 from proof initial-ex-abstract-initial (previous step)

m0 +1 = # pebs, ubi-io-initial(pebs), ∀ pebs. m0 = # pebs ∧ ubi-io-initial(pebs) → (∃ apebs. abs(pebs, apebs) ∧ ubi-io-initial(apebs)) ⊦ ∃ apebs. abs(pebs, apebs) ∧ ubi-io-initial(apebs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)