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Proof step 16 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
x1 + a ' + b ' + y + c ' + z = (y0 + c ' + a ') + z0, ∀ a, b, y, c, z, x0, z0. x1 + a ' + b ' + y + c ' + z = x0 + c ' + a ' + z0 → a = b ∨ a = c ∨ a ∈ y ∨ a ∈ x1 ∨ a ∈ z ∨ c ∈ x1 ∨ c ∈ z ∨ dups(x1 + z) ⊦ a = c, a = b, c = a0, a = a0, c ∈ z, c ∈ x1, a ∈ z, a ∈ x1, a ∈ y, dups(x1 + z), a0 ∈ z, a0 ∈ x1

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c-08: ⊦ (a ' + x).first = a; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c-07: ⊦ (a ' + x).rest = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list