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Proof step 5 from proof lastn-neq-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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

Rule argument

Additional information

Used simplifier rules: in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%Last: ⊦ x + a = x + a '; from specification list