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Proof step 2 from proof inapp-01 (previous step)

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

Rule argument

Additional information

Used simplifier rules: ⊦ 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