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Proof step 17 from proof eq-01 (previous step)

x1 + a ' + b ' + y + c ' + z = (x0 + a ' + c ').tail + z0, x0 = a0 + y0, (x0 + a ').head ∈ x0, ∀ a, b, y, c, z, x0, z0. x1 + a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 → a = b ∨ a = c ∨ b = c ∨ dups(x1 + z) ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ⊦ a = a0, a = b, a = c, b = c, c = a0, x0 = [], dups(x1 + z), a0 ∈ x1, a0 ∈ z, a ∈ y, a ∈ x1, a ∈ z, c ∈ x1, c ∈ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
x1 + a ' + b ' + y + c ' + z = (y0 + a ' + c ') + z0, ∀ a, b, y, c, z, x0, z0. x1 + a ' + b ' + y + c ' + z = x0 + a ' + c ' + z0 → a = b ∨ a = c ∨ b = c ∨ dups(x1 + z) ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ⊦ a = c, a = b, a = a0, b = c, c = a0, dups(x1 + z), c ∈ z, c ∈ x1, a ∈ z, a ∈ x1, a ∈ y, a0 ∈ z, a0 ∈ x1

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ (a ' + x).head = a; (s)% ⊦ a ' + x ≠ []; (s)% ⊦ (a ' + x).tail = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)