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Proof step 6 from proof rev-rest (previous step)

rev(x) = a0 ' + y1, rev(x) + y = a ' + y0, a0 ∈ x, a ∈ x ∨ a ∈ y ⊦ y0 = y1 + y, x = []

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: x

Additional information

Used simplifier rules: rev-r: ⊦ rev(a ' + x) = rev(x) + a;%list freely generated by [], +;;%c: ⊦ a + x = a ' + x; from specification list%Last: ⊦ x + a = x + a '; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; 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%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list