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Proof step 5 from proof app-06 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x0 + a ' = y0 + b ', b = a0 ∨ b ∈ x0 ∨ b = a, a = a0 ∨ a ∈ y0 ∨ a = b, ∀ x, a, b. x + a ' = y0 + b ' ∧ (a ∈ y0 ∨ a = b) ∧ (b ∈ x ∨ b = a) → x = y0 ∧ a = b, a ∈ y0 ∨ a = b, b ∈ x0 ∨ b = a ⊦ x0 = y0 ∧ a = b

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%in-15: ⊦ x + a ' = y → a ∈ y;%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y);%in-17: ⊦ a ∈ b ' ↔ a = b;