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Proof step 6 from proof eq-01 (previous step)

x0 ≠ [], a = x0.first ∧ b ' + y + c ' + z = x0.rest + a ' + c ' + z0 ⊦ a = c, a = b, b = c, dups(z), c ∈ z, a ∈ z, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
b ' + y + c ' + z = x0.rest + x0.first ' + c ' + z0, b ∈ x0.rest ⊦ x0 = [], b = c, x0.first = b, x0.first = c, x0.first ∈ y, x0.first ∈ z, c ∈ z, dups(z)

Rule argument

Additional information

Used simplifier rules: inapp-02: ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list