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Proof step 9 from proof unequal (previous step)

a1 = a, # x0 = # y0, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ a1 ' + x0 = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a1 ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# x0 = # y0, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x0 = y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Rule argument

Additional information

Used simplifier rules: app-08: ⊦ a ' + x = a ' + y ↔ x = y;%a: ⊦ (x + y) + z = x + y + z;