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Proof step 6 from proof lastn-neq-01 (previous step)

lastn(n0, x.butlast) + x.last ' = b ' + y, n0 < # x, a ∈ y, ∀ x, a, y, b. n0 ≤ # x ∧ a ∈ y ∧ ¬ a ∈ x → lastn(n0, x) ≠ b ' + y, 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 inst: name: last-elim seq: x ≠ [] ⊦ a = x.last ∧ y = x.butlast ↔ x = y + a ' subst: [x, a, y] → [x, a0, y0] flags: []

Additional information

Used simplifier rules: lastN-in-01: ⊦ ¬ a ∈ x ∧ n ≤ # x → ¬ a ∈ lastn(n, x);%e-02: ⊦ x + a ' ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%len-02: ⊦ # a ' = 0 +1; from specification list%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%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