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Proof step 5 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 6:
# x = # y0 + 1, x = a1 + x0, ∀ 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 [], +;;% ⊦ # [] = 0; (s)% ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic