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Proof step 12 from proof fun-value-and-arg-reduces-imp-app-reduces (previous step)

t => t', value(t1) ⊦ t1 _ t => t1 _ t'

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ t => t' ↔ (∃ C. (∃ v1, v2. t = C[K _ v1 _ v2] ∧ t' = C[v1] ∧ value(v1) ∧ value(v2) ∧ context(C)) ∨ (∃ v1, v2, v3. t = C[S _ v1 _ v2 _ v3] ∧ t' = C[v1 _ v3 _ (v2 _ v3)] ∧ value(v1) ∧ value(v2) ∧ value(v3) ∧ context(C)))
Proof step 11:
∃ C. (∃ v1, v2. t = C[K _ v1 _ v2] ∧ t' = C[v1] ∧ value(v1) ∧ value(v2) ∧ context(C)) ∨ (∃ v1, v2, v3. t = C[S _ v1 _ v2 _ v3] ∧ t' = C[v1 _ v3 _ (v2 _ v3)] ∧ value(v1) ∧ value(v2) ∧ value(v3) ∧ context(C)), value(t1) ⊦ t1 _ t => t1 _ t'

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (t', t)]]