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Proof step 43 from proof termination-RED1 (previous step)

〈RED1(t4; tn2)〉 tn2 = none, 〈RED1(t3; tn1)〉 tn1 = none, ∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t3 .1 .1 = S, t3 ≠ K, t3 .1 ≠ S, value(t3 _ t4), value(t3), ap(t3), ap(t3 .1), ¬ hasS(t2), ¬ hasS(t3), ¬ hasS(t4), ¬ (t3 .1 = K ∧ value(t3 .2) ∧ value(t4)), ¬ (value(t3 .1 .2) ∧ value(t3 .2) ∧ value(t4)) ⊦ 〈RED1(t2; tn0)〉〈t1 := t3 _ t4 ; if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 42:
〈RED1(t4; tn2)〉 tn2 = none, 〈RED1(t3; tn1)〉 tn1 = none, ∀ t0. #(t0) < #(t3 _ t4 _ t2) → (¬ hasS(t0) → 〈RED1(t0; tn)〉 (tn ≠ none → ¬ hasS(tn) ∧ #(tn) < #(t0))), t5 .1 = S, t3 ≠ K, t5 ≠ S, value(t3 _ t4), value(t3), ap(t3), ap(t5), ¬ hasS(t2), ¬ hasS(t3), ¬ hasS(t4), ¬ (t5 = K ∧ value(t6) ∧ value(t4)), ¬ (value(t5 .2) ∧ value(t6) ∧ value(t4)), t3 = t5 _ t6 ⊦ 〈RED1(t2; tn0)〉〈t1 := t3 _ t4 ; if tn0 ≠ none then tn0 := some(t1 _ tn0 .t)〉 (tn0 ≠ none → ¬ hasS(tn0) ∧ ¬ #(t1) + #(t2) < #(tn0))

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] → [t3, t, t5] 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 : (t3, t, t5)]]