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Proof step 6 from proof adjoin (previous step)

¬ a ∈ x ⊦ perm(x, y) → perm(a + x, y ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
perm(x, y), a ∈ y ++ a, perm(y, x) ⊦ perm(x, (y ++ a) -1l a), a ∈ x

Rule argument

Additional information

Used simplifier rules: del-01: ⊦ perm(a ' + x, y) ↔ ¬ ¬(a ∈ y ∧ perm(x, y -1l a));%sym-01: ⊦ perm(x, y) → perm(y, x);%trans: ⊦ perm(x, y) ∧ perm(y, z) → perm(x, z);%sym-02: ⊦ ¬ perm(x, y) → ¬ perm(y, x);%ref: ⊦ perm(x, x);%sym-02: ⊦ ¬ perm(x, y) → ¬ perm(y, x);%in-08: ⊦ a ∈ x ++ b ↔ a = b ∨ a ∈ x; from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list