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Proof step 19 from proof initial-ex-abstract-initial (previous step)

m0 + 1 = # pebs, n < m0, ∀ n. n < # pebs → ubi-io-peb-initial(pebs[n]) ⊦ ubi-io-peb-initial(pebs[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
m0 + 1 = # pebs, n < m0, ∀ n0. ¬ m0 < n0 → ubi-io-peb-initial(pebs[n0]) ⊦ ubi-io-peb-initial(pebs[n])

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)