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Proof step 10 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)

User applied rule all left

And obtained the following new goals

Proof step 11:
m0 + 1 = # pebs, ubi-io-initial(pebs), m0 = # subarray(pebs, 0, m0) ∧ ubi-io-initial(subarray(pebs, 0, m0)) → (∃ apebs. abs(subarray(pebs, 0, m0), apebs) ∧ ubi-io-initial(apebs)) ⊦ ∃ apebs. abs(pebs, apebs) ∧ ubi-io-initial(apebs)

Rule argument

pos: left 3 terms: subarray(pebs, 0, m0) abort: F discard: T computed: F

Additional information

Used terms: subarray(pebs, 0, m0)