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Proof step 10 from proof ks-TOPRED (previous step)

ks(n) .1 = K _ t1, value(ks(n) .2), value(t1), ap(ks(n)) ⊦ n = 2 ∧ ks(2) .1 .2 = K

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
ks(n) = t0 _ t2, t0 = K _ t1, value(t2), value(t1), ap(ks(n)) ⊦ n = 2 ∧ ks(2) .1 .2 = K

Rule argument

[x0lemmaarg: spec-name: term instance-name: lemma-name: elim antecedent: succedent: ap(t) → (t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1) substlist: [t, t0, t1] → [ks(n), t, t0] add preconditions?: F is rewrite lemma?: F current spec?: F rotate lemma?: T replace all?: T paths: () dummy: () ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ ap(t) → ( t0 = t .1 ∧ t1 = t .2 ↔ t = t0 _ t1)])], [EXTRATERMS : (ks(n), t, t0)]]