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Proof step 12 from proof unequal (previous step)

# z0 = # z1, ∀ x. # x = # z + # z1 + 1 → x = z + b ' + z1 ∨ (∃ z2, a, b0, z0, z3. a ≠ b0 ∧ x = z2 + a ' + z0 ∧ z + b ' + z1 = z2 + b0 ' + z3) ⊦ a0 = b, a0 ' + z0 = b ' + z1, ∃ z2, a1, b0, z4, z3. a1 ≠ b0 ∧ a ' + z + a0 ' + z0 = z2 + a1 ' + z4 ∧ a ' + z + b ' + z1 = z2 + b0 ' + z3

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: a ' + z, a0, b, z0, z1

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; Used terms: a ' + z, a0, b, z0, z1