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Proof step 75 from proof infinite-append-butlast (previous step)

m1 + 1 = ii(n0) + # lf(n0).tail, (ii(n0) - 1) + # lf(n0) = m0 + 1, ii(0) = 0, ii(n0 + 1) = m1 + 1, (σ from ii(n0)) to # lf(n0).tail = ⌜lf(n0).tail⌝ , ⌜m1⌝ < # σ, (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).head, ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = ⌜(n = 0 ⊃ lf(0);lf(n).tail)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ (ii(n0) - 1) + # lf(n0) = 0, m1 = m0, n0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 76:
m1 + 1 = ii(n0) + # lf(n0).tail, (ii(n0) - 1) + # lf(n0) = m0 + 1, ii(0) = 0, ii(n0 + 1) = m1 + 1, (σ from ii(n0)) to # lf(n0).tail = ⌜lf(n0).tail⌝ , ⌜m1⌝ < # σ, (finp(σ) ⊃ # σ = ⌜ii(n)⌝ ∧ (∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).tail) = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).head, ∀ n. lf(n).last = lf(n + 1).head ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).tail) ∧ ¬ # σ < ⌜ii(n + 1)⌝ ∧ (σ from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).tail) = ⌜(n = 0 ⊃ lf(0);lf(n).tail)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ m1 = m0, n0 = 0

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 + 1 ≠ 0; (s)