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Proof step 87 from proof infinite-append-butlast (previous step)

ii(0) = 0, ∀ n. n ≠ 0 → ii(n) ≠ 0, ii(0 + 1) = ii(0) + # (0 = 0 ⊃ lf(0);lf(0).rest) ∧ ¬ # σ < ⌜ii(0 + 1)⌝ ∧ (σ from ii(0)) to # (0 = 0 ⊃ lf(0);lf(0).rest) = ⌜(0 = 0 ⊃ lf(0);lf(0).rest)⌝ , ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = ⌜(n = 0 ⊃ lf(0);lf(n).rest)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)) ⊦ σ[ii(1) - 1] = lf(1).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 88:
ii(0) = 0, σ to ii(1) = ⌜lf(0)⌝ , ii(1) = # lf(0), (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = ⌜(n = 0 ⊃ lf(0);lf(n).rest)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[ii(1) - 1] = lf(1).first, # σ < ⌜ii(1)⌝

Rule argument

Additional information

Used simplifier rules: add-base: ⊦ n + 0 = n; from specification nat-basic2%from-zero: ⊦ σ from 0 = σ; from specification stream%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2