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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%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% ⊦ x + a = x + a '; (s)