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Proof step 3 from proof del1-empty (previous step)

∀ a. x0 -1l a = [] → x0 = [] ∨ x0 = a ', (a0 + x0) -1l a = [] ⊦ a0 + x0 = [], a0 + x0 = a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
(a0 ' + x0) -1l a = [], ∀ a. x0 -1l a = [] → x0 = [] ∨ x0 = a ' ⊦ a0 ' + x0 = a '

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list