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Proof step 4 from proof app-06 (previous step)

x + a ' = (a0 ' + y0) + b ', b ∈ x ∨ b = a, ∀ x, a, b. x + a ' = y0 + b ' ∧ (a ∈ y0 ∨ a = b) ∧ (b ∈ x ∨ b = a) → x = y0 ∧ a = b, a = a0 ∨ a ∈ y0 ∨ a = b, a0 ∈ x ∨ a0 = a ⊦ x = a0 ' + y0 ∧ a = b

User applied rule constructor cut

And obtained the following new goals

Proof step 5:
x = a1 + x0, x + a ' = (a0 ' + y0) + b ', b ∈ x ∨ b = a, ∀ x, a, b. x + a ' = y0 + b ' ∧ (a ∈ y0 ∨ a = b) ∧ (b ∈ x ∨ b = a) → x = y0 ∧ a = b, a = a0 ∨ a ∈ y0 ∨ a = b, a0 ∈ x ∨ a0 = a ⊦ x = a0 ' + y0 ∧ a = b

Rule argument

var: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-03: ⊦ a ' = b ' + x ↔ a = b ∧ x = [];%e-02: ⊦ x + a ' ≠ [];%list freely generated by [], +;;