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Proof step 5 from proof lr-03 (previous step)

x = a + y0 ⊦ # y0 = # y.tail ↔ # x = # y, x = [], y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ # y0 = # y.tail ↔ # y0 + 1 = # y, y = []

Rule argument

Additional information

Used simplifier rules: disj-1: ⊦ [] ≠ a + x;%size-02: ⊦ #(a + x) = (# x)+1;% ⊦ m +1 = m + 1; (s)