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Proof step 4 from proof len-06 (previous step)

# x = # z → (x + y = z + y0 ↔ x = z ∧ y = y0), # y = # y0, x + y = z + y0 ⊦ x = z ∧ y = y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# y = # y0, x + y = z + y0 ⊦ # x = # z

Rule argument

Additional information

Used simplifier rules: length-neq: ⊦ # x ≠ # y → x ≠ y; from specification list-data