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Proof step 11 from proof termination-TOPRED (previous step)

t .1 .1 .1 = S, ap(t), ap(t .1), ap(t .1 .1), value(t .1 .1 .2), value(t .1 .2), value(t .2) ⊦ t .1 .1 = K, hasS(t), ¬ hasS(t .1 .1 .2 _ t .2 _ (t .1 .2 _ t .2)) ∧ 3 + #(t .1 .1 .2) + #(t .1 .2) + 2 * #(t .2) < #(t)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
t = t0 _ t1, t0 .1 .1 = S, ap(t), ap(t0), ap(t0 .1), value(t0 .1 .2), value(t0 .2), value(t1) ⊦ t0 .1 = K, hasS(t), ¬ hasS(t0 .1 .2 _ t1 _ (t0 .2 _ t1)) ∧ 3 + #(t0 .1 .2) + #(t0 .2) + 2 * #(t1) < #(t)

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] → [t, t0, t1] 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 : (t, t0, t1)]]