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Proof step 67 from proof infinite-append-butlast (previous step)

(σ from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = ⌜(n0 = 0 ⊃ lf(0);lf(n0).rest)⌝ , m1 + 1 = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).rest), ii(n0 + 1) = m1 + 1, ii(0) = 0, ⌜m1⌝ < # σ, ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ 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)) ⊦ m1 = (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%from-zero: ⊦ σ from 0 = σ; from specification stream%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat