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Proof step 27 from proof infinite-append-butlast (previous step)

ii(n0) = m0 + 1, ii(0) = 0, (σ from m0 + 1) to # lf(n0).rest = ⌜lf(n0).rest⌝ , ii(n0 + 1) = (# lf(n0).rest + m0) + 1, m0 + # lf(n0) - 1 = # lf(n0).rest + m0, ⌜m0⌝ < # σ, ⌜# lf(n0).rest⌝ + ⌜m0⌝ < # 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)) ⊦ ⌜σ[m0] '⌝ + (σ from m0 + 1) to # lf(n0) = ⌜lf(n0)⌝

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: name: elim-1 seq: ⊦ x ≠ [] → (a = x.first ∧ x0 = x.rest ↔ x = a + x0) subst: [x, a, x0] → [lf(n0), a, x] flags:

Additional information

Used simplifier rules: size-01: ⊦ # [] = 0; from specification list-data%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%to-zero: ⊦ σ to 0 = []; from specification stream%empty: ⊦ ⌜[]⌝ = []; from specification stream% ⊦ x ≠ [] → (a = x.first ∧ x0 = x.rest ↔ x = a + x0); (elim) Used terms: lf(n0), a, x