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Proof step 10 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) ⊦ x0 = y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

User applied rule all left

And obtained the following new goals

Proof step 11:
# x0 = # y0, # x0 = # y0 → x0 = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x0 = z + a ' + z0 ∧ y0 = z + b ' + z1), ∀ 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

pos: left 2 terms: x0 abort: F discard: F computed: T

Additional information

Used terms: x0