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Proof step 1 from proof unequal (conclusion)

# x = # y, x ≠ y ⊦ ∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y = z + b ' + z1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# x = # y ⊦ x = y, ∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y = z + b ' + z1

Rule argument

Additional information