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Proof step 5 from proof neq-01 (previous step)

∀ a0, y0, c, z, a, x. (a0 ' + y0) + c ' + z = y + a ' + x ∧ a0 ∈ y ∧ (∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0) → a = a0 ∨ a = c ∨ a ∈ y0 ∨ a ∈ z, (a0 ' + y0) + c ' + z = (a1 + y) + a ' + x, a0 ∈ a1 + y, ∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0 ⊦ a = a0, a = c, a ∈ y0, a ∈ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
y0 + c ' + z = y + a ' + x, ∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0, ∀ a0, y0, c, z, a, x. (a0 ' + y0) + c ' + z = y + a ' + x ∧ a0 ∈ y ∧ (∀ c, z, y1, a, x. y0 + c ' + z = y1 + a ' + x → a = c ∨ a ∈ z ∨ a ∈ y0) → a = a0 ∨ a = c ∨ a ∈ y0 ∨ a ∈ z ⊦ a = c, a = a0, a ∈ z, a ∈ y0

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);