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Proof step 3 from proof rev-rest (previous step)

rev(x) + y = a + y0 ∧ rev(x) + y ≠ [] ⊦ y0 = rev(x).rest + y, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
rev(x) + y = a ' + y0, a ∈ x ∨ a ∈ y ⊦ x = [], y0 = rev(x).rest + y

Rule argument

Additional information

Used simplifier rules: rev-in: ⊦ a ∈ rev(x) ↔ a ∈ x;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%c: ⊦ a + x = a ' + x; from specification list