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Proof step 46 from proof diag-theorem-butlast (previous step)

lf(m + 1) = a0 + x0 ∧ lf(m + 1) ≠ [], # nextl(0, a) = ii(1) + 1, lf(n0 + 1) = nextl(n0, lf(n0).last), x to ii(1) + 1 = nextl(0, a), ii(0) = 0, lf(0) = a ', lf(1) = nextl(0, a), nextl(0, a) = a ' + y, ii(n0 + 1) = ii(n0) + # lf(n0 + 1) - 1, (x from ii(n0)) to # lf(n0 + 1) = lf(n0 + 1), 1 + ii(n) = # x, lf(m + 1) = nextl(m, x[ii(n)]), x[ii(n)] = lf(m).last, x0 = [], ii(m) = ii(n), a ∈ nextl(0, a), ii(1) ≤ ii(n), ii(n0 + 1) ≤ ii(n), n ≤ m, ∀ n1. ii(n1 + 1) = ii(n1) + # lf(n1 + 1) - 1 ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # lf(n1 + 1) = lf(n1 + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ∧ nextl(n, a).first = a ⊦ nextl(0, a) = [], lf(n0) = [], x = [], lf(m) = [], lf(m + 1) = x[ii(n)] '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
lf(m + 1) = a0 ', ii(m) = ii(n), x[ii(n)] = lf(m).last, a0 ' = nextl(m, x[ii(n)]), 1 + ii(n) = # x, (x from ii(n0)) to # lf(n0 + 1) = lf(n0 + 1), ii(n0 + 1) = ii(n0) + # lf(n0 + 1) - 1, nextl(0, a) = a ' + y, lf(1) = nextl(0, a), lf(0) = a ', ii(0) = 0, x to ii(1) + 1 = nextl(0, a), lf(n0 + 1) = nextl(n0, lf(n0).last), # nextl(0, a) = ii(1) + 1, n ≤ m, ii(n0 + 1) ≤ ii(n), ii(1) ≤ ii(n), a ∈ nextl(0, a), ∀ n, a. nextl(n, a) ≠ [] ∧ nextl(n, a).first = a, ∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n1. ii(n1 + 1) = ii(n1) + # lf(n1 + 1) - 1 ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # lf(n1 + 1) = lf(n1 + 1) ⊦ lf(m) = [], x = [], lf(n0) = [], nextl(0, a) = [], a0 = x[ii(n)]

Rule argument

Additional information

Used simplifier rules: c-04: ⊦ a ' = b ' ↔ a = b; from specification list%e: ⊦ a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app: ⊦ x + [] = x; from specification list%c: ⊦ a + x = a ' + x; from specification list