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Proof step 6 from proof unequal (previous step)

# 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 weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
# x0 = # y0, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ a ' = [], a1 ' + x0 = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a1 ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Rule argument

fma: a ' = []

Additional information

Used simplifier rules: e: ⊦ a ' ≠ []; Used formulas: a ' = []