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Proof step 42 from proof initial-ex-abstract-initial (previous step)

# apebs + 1 = # pebs, pebs[# apebs] = mkpeb(buf, m, true), n < # apebs, n < # mkapebs(# pebs), ubi-io-initial(pebs), ubi-io-initial(apebs), ∀ n0. n0 < # apebs → abs-peb(pebs[n0], apebs[n0]) ⊦ n = # apebs, abs-peb(pebs[n], (n < 0 ∨ 0 + # apebs ≤ n ⊃ mkapebs(# pebs)[n];apebs[n + 0 - 0]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
# apebs + 1 = # pebs, pebs[# apebs] = mkpeb(buf, m, true), n < # pebs, n < # apebs, ubi-io-initial(pebs), ubi-io-initial(apebs), ∀ n0. n0 < # apebs → abs-peb(pebs[n0], apebs[n0]) ⊦ n = # apebs, abs-peb(pebs[n], apebs[n])

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ n → (m < n + 1 ↔ m < n); (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m - 0 = m; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ ¬ m < 0; (s)% ⊦ # mkapebs(n) = n; (s)