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

∀ a0, x1, a, y, b, z. (a0 ' + x1) + a ' + y = x + b ' + z ∧ a0 ∈ x ∧ (∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → a ∈ z ∨ b ∈ y ∨ x1 = x0 ∧ a = b ∧ y = z) → b ∈ y ∨ a ∈ z ∨ a0 ' + x1 = x ∧ a = b ∧ y = z, (a0 ' + x1) + a ' + y = (a1 + x) + b ' + z, a0 ∈ a1 + x, ∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → a ∈ z ∨ b ∈ y ∨ x1 = x0 ∧ a = b ∧ y = z ⊦ b ∈ y, a ∈ z, a0 ' + x1 = a1 + x ∧ a = b ∧ y = z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x1 + a ' + y = x + b ' + z, ∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → a ∈ z ∨ b ∈ y ∨ x1 = x0 ∧ a = b ∧ y = z, ∀ a0, x1, a, y, b, z. (a0 ' + x1) + a ' + y = x + b ' + z ∧ a0 ∈ x ∧ (∀ a, y, x0, b, z. x1 + a ' + y = x0 + b ' + z → a ∈ z ∨ b ∈ y ∨ x1 = x0 ∧ a = b ∧ y = z) → b ∈ y ∨ a ∈ z ∨ a0 ' + x1 = x ∧ a = b ∧ y = z ⊦ a ∈ z, b ∈ y, x1 = x ∧ a = b ∧ y = z

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);