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Proof step 3 from proof cons-01 (previous step)

∀ x, a. ¬(disj(x, y0) ∧ a ∈ y0 ∧ disj(a ' + x, y0)), disj(x, a0 + y0), a ∈ a0 + y0, disj(a ' + x, a0 + y0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
disj(x, y0), disj(a ' + x, y0), a ∈ y0, ∀ x0, a1. ¬ disj(x0, y0) ∨ ¬ a1 ∈ y0 ∨ ¬ disj(a1 ' + x0, y0) ⊦ a0 = a, a0 ∈ x

Rule argument

Additional information

Used simplifier rules: app-02: ⊦ disj(x, y + z) ↔ disj(x, y) ∧ disj(x, z);%one-01: ⊦ disj(x, a ') ↔ ¬ a ∈ x;%in-10: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list