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Proof step 3 from proof key-cons-app (previous step)

∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons((nd + ndx) + ndx1) ↔ key-cons(nd + ndx) ∧ key-cons(ndx1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
key-cons(nd.key, nd), ∀ ndx1. key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1) ⊦ key-cons(ndx + ndx1) ↔ key-cons(ndx) ∧ key-cons(ndx1)

Rule argument

Additional information

Used simplifier rules: key-cons-rec: ⊦ key-cons(nd ' + ndx) ↔ ¬ ¬(key-cons(nd.key, nd) ∧ key-cons(ndx));%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ nd + ndx = nd ' + ndx; (s)