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Proof step 7 from proof lastn-neq-01 (previous step)

lastn(n0, x.butlast) + x.last ' = b ' + y, n0 < # x, a ∈ y, ∀ x, b, y, a. lastn(n0, x) = b ' + y ∧ a ∈ y ∧ n0 ≤ # x → a ∈ x, x.last = b ∨ x.last ∈ y, b ∈ lastn(n0, x.butlast) ∨ b = x.last ⊦ x = [], a ∈ x

User applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-last name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x, a0, y0]

Additional information

Used simplifier rules: lastN-in-01: ⊦ ¬ a ∈ x ∧ n ≤ # x → ¬ a ∈ lastn(n, x);% ⊦ x + a ' ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ #(x + y) = # x + # y; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)%inapp-13: ⊦ a ' + x = y + b ' ∧ c ∈ x ∧ c ≠ b → c ∈ y; (forward) from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a '; (elim) Used terms: x, a0, y0