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Proof step 2 from proof lastN-rest-01 (previous step)

# x = n + 1 ⊦ lastn(n, x) = x.rest

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y] flags:

Additional information

Used simplifier rules: lastN-append: ⊦ n = # y → lastn(n, x + y) = y;%c: ⊦ a + x = a ' + x; from specification list%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: x, a, y