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Proof step 4 from proof unequal (previous step)

# x = # y0 + 1, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
x = a1 + x0, # x = # y0 + 1, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Rule argument

var: x

Additional information

Used simplifier rules: list freely generated by [], +;;%size-01: ⊦ # [] = 0; from specification list-data%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2