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Proof step 5 from proof insperm-03 (previous step)

perm(a ' + x, ins≤(a, x)), perm(ins≤(a, x), a ' + x) ⊦ perm(ins≤(a, x), y) ↔ perm(a ' + x, y)

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: trans: ⊦ perm(x, y) ∧ perm(y, z) → perm(x, z); (forward) from specification list-perm%sym-01: ⊦ perm(x, y) → perm(y, x); (forward) from specification list-perm%sym-02: ⊦ ¬ perm(x, y) → ¬ perm(y, x); (forward) from specification list-perm