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Proof step 31 from proof TOPRED-lem (previous step)

t =>top t0 ⊦ ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S, ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 28:
⊦ t =>top t' ↔ (∃ v1, v2. t = K _ v1 _ v2 ∧ t' = v1 ∧ value(v1) ∧ value(v2)) ∨ (∃ v1, v2, v3. t = S _ v1 _ v2 _ v3 ∧ t' = v1 _ v3 _ (v2 _ v3) ∧ value(v1) ∧ value(v2) ∧ value(v3))
Proof step 30:
(∃ v1, v2. t = K _ v1 _ v2 ∧ t0 = v1 ∧ value(v1) ∧ value(v2)) ∨ (∃ v1, v2, v3. t = S _ v1 _ v2 _ v3 ∧ t0 = v1 _ v3 _ (v2 _ v3) ∧ value(v1) ∧ value(v2) ∧ value(v3)) ⊦ ap(t) ∧ ap(t .1) ∧ ap(t .1 .1) ∧ t .1 .1 .1 = S, ap(t) ∧ ap(t .1) ∧ t .1 .1 = K ∧ value(t .1 .2) ∧ value(t .2)

Rule argument

[x0lemmaarg: spec-name: reduce instance-name: lemma-name: topred-def antecedent: succedent: t =>top t' ↔ (∃ v1, v2. t = K _ v1 _ v2 ∧ t' = v1 ∧ value(v1) ∧ value(v2)) ∨ (∃ v1, v2, v3. t = S _ v1 _ v2 _ v3 ∧ t' = v1 _ v3 _ (v2 _ v3) ∧ value(v1) ∧ value(v2) ∧ value(v3)) substlist: [t', t] → [t0, t] add preconditions?: F is rewrite lemma?: T current spec?: F rotate lemma?: F replace all?: T paths: ((0, 0, 0)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (t0, t)]]