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Proof step 3 from proof l (previous step)

# x = m + 1 ∧ # x ≠ 0, a ∈ x ⊦ #(x -1l a) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x = m + 1, a ∈ x ⊦ #(x -1l a) = m, x = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data