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Proof step 2 from proof inapp (previous step)

a ' + x + b ' = y + c ' + c ' + z ⊦ a = b, dups(x), b ∈ x, a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ' + x + b ' = y + c ' + c ' + z, a ∈ y, b ∈ y ∨ b = c ∨ b ∈ z ⊦ a = b, dups(x), a ∈ x, b ∈ x

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b); (s)%inapp-11: ⊦ x + y = z + a ' ∧ b ∈ x ∧ a ≠ b → b ∈ z; (forward) from specification list% ⊦ c ' + x + b ' = y + a ' + a ' + z ∧ a ≠ b → a ∈ x; (s)% ⊦ ¬ a ∈ y → (y + c ' + z = a ' + z0 ↔ y = [] ∧ c = a ∧ z = z0); (ws)% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) 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