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Proof step 1 from proof inapp-01 (conclusion)

⊦ a ' + x + b ' = y + c ' + a0 ' + z ∧ a ≠ b ∧ ¬ a ∈ x ∧ ¬ b ∈ x ∧ c ∈ z → dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ' + x + b ' = y + c ' + a0 ' + z, c ∈ z, a ∈ y ∨ a = a0 ∨ a ∈ z, b ∈ y ∨ b = a0 ∨ b ∈ z ⊦ a = b, a ∈ x, b ∈ x, dups(x)

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list