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Proof step 6 from proof inapp (previous step)

a ' + x + b ' = y + c ' + c ' + z, a ∈ y, b ∈ y ∨ b = c ∨ b ∈ z, z ≠ [] → z = y0 + a0 ' ⊦ z = [], a = c, a = b, dups(x), b ∈ x, a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ' + x = y + c ' + c ' + y0, a ∈ y ⊦ a = c, a = b, dups(x), b ∈ x, a ∈ x

Rule argument

Additional information

Used simplifier rules: in-11: ⊦ a ∈ x → a ∈ x + y; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%e-02: ⊦ x + a ' ≠ []; from specification list%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list