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Proof step 4 from proof inapp-01 (previous step)

z ≠ [] → (∃ y, a. z = y + a '), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ' + x = y + c ' + a0 ' + y0, c ∈ y0, a ∈ y0 ∨ a ∈ y, a ∈ y ∨ a ∈ y0 ⊦ a = a0, a = a1, c = a1, dups(x), a1 ∈ x, a ∈ x

Rule argument

Additional information

Used simplifier rules: neq-02: ⊦ ¬ a ∈ z ∧ a ≠ c → c ' + z ≠ y1 + a ' + x; from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%inapp-04: ⊦ ¬ a ∈ x → (a ' + x = y + a ' + z ↔ y = [] ∧ z = x); 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%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%ni: ⊦ x + a ' = y + z ∧ ¬ a ∈ z → z = []; (forward) from specification list%in-empty: ⊦ ¬ 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%app: ⊦ x + [] = x; from specification list