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Proof step 5 from proof postfix-rec (previous step)

x = y0 + a ' ⊦ x ⊒ y ↔ a = y.last ∧ y0 ⊒ y.butlast, x = [], y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ y0 + a ' ⊒ y ↔ a = y.last ∧ y0 ⊒ y.butlast, y = []

Rule argument

Additional information

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