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Proof step 5 from proof inapp-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ' + x = y + c ' + a0 ' + y0, c ∈ y0, a0 ∈ x, c ∈ x ∨ c = a, a ∈ y ∨ a ∈ y0, c = a ∨ c ∈ x ⊦ a = b, c = b, dups(x), a ∈ x, b ∈ x

Rule argument

Additional information

Used simplifier rules: in-13: ⊦ x = y + a ' + z → a ∈ x; (forward) from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list% ⊦ ¬ b ∈ x → c ' + x ≠ y + a ' + b ' + z; (s)% ⊦ ¬ a ∈ x → (a ' + x = y + a ' + z ↔ y = [] ∧ z = x); (ws)% ⊦ ¬ a ∈ z ∧ a ≠ c → c ' + z ≠ y1 + a ' + x; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x + a ' ≠ []; (s)%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []; (forward) from specification list% ⊦ ¬ a ∈ []; (s)% ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)