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Proof step 7 from proof fun-value-and-arg-reduces-imp-app-reduces (previous step)

value(t1), t = C[K _ v _ v0] ∧ t' = C[v] ∧ value(v) ∧ value(v0) ∧ context(C) ∨ t = C[S _ v1 _ v2 _ v3] ∧ t' = C[v1 _ v3 _ (v2 _ v3)] ∧ value(v1) ∧ value(v2) ∧ value(v3) ∧ context(C) ⊦ ∃ v4, v5. t1 _ t = t1 _ C[K _ v4 _ v5] ∧ t1 _ t' = t1 _ C[v4] ∧ value(v4) ∧ value(v5) ∧ context(C), ∃ v4, v5, v6. t1 _ t = t1 _ C[S _ v4 _ v5 _ v6] ∧ t1 _ t' = t1 _ C[v4 _ v6 _ (v5 _ v6)] ∧ value(v4) ∧ value(v5) ∧ value(v6) ∧ context(C)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
t = C[K _ v _ v0] ∧ t' = C[v] ∧ value(v) ∧ value(v0) ∧ context(C), value(t1) ⊦ ∃ v4, v5. t1 _ t = t1 _ C[K _ v4 _ v5] ∧ t1 _ t' = t1 _ C[v4] ∧ value(v4) ∧ value(v5) ∧ context(C), ∃ v4, v5, v6. t1 _ t = t1 _ C[S _ v4 _ v5 _ v6] ∧ t1 _ t' = t1 _ C[v4 _ v6 _ (v5 _ v6)] ∧ value(v4) ∧ value(v5) ∧ value(v6) ∧ context(C)
Proof step 6:
t = C[S _ v1 _ v2 _ v3] ∧ t' = C[v1 _ v3 _ (v2 _ v3)] ∧ value(v1) ∧ value(v2) ∧ value(v3) ∧ context(C), value(t1) ⊦ ∃ v4, v5. t1 _ t = t1 _ C[K _ v4 _ v5] ∧ t1 _ t' = t1 _ C[v4] ∧ value(v4) ∧ value(v5) ∧ context(C), ∃ v4, v5, v6. t1 _ t = t1 _ C[S _ v4 _ v5 _ v6] ∧ t1 _ t' = t1 _ C[v4 _ v6 _ (v5 _ v6)] ∧ value(v4) ∧ value(v5) ∧ value(v6) ∧ context(C)

Rule argument

[CASEDARG : pos: left 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]