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Proof step 29 from proof subst-SK-equal (previous step)

C0 = O, C1[S _ v3 _ v4 _ v5] _ t = C0[K _ v _ v0], value(v3), value(v4), value(v5), value(v), value(v0), context(C0), context(C1), ∀ v1, v2, v6, C, v7, v8. C1[S _ v1 _ v2 _ v6] = C[K _ v7 _ v8] ∧ value(v1) ∧ value(v2) ∧ value(v6) ∧ value(v7) ∧ value(v8) ∧ context(C) → C1[v1 _ v6 _ (v2 _ v6)] = C[v7] ⊦ C1[v3 _ v5 _ (v4 _ v5)] _ t = C0[v]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ O[t] = t
Proof step 28:
C1[S _ v3 _ v4 _ v5] = K _ v, context(C1), value(v0), value(v), value(v5), value(v4), value(v3), ∀ v1, v2, v6, C, v7, v8. C1[S _ v1 _ v2 _ v6] = C[K _ v7 _ v8] ∧ value(v1) ∧ value(v2) ∧ value(v6) ∧ value(v7) ∧ value(v8) ∧ context(C) → C1[v1 _ v6 _ (v2 _ v6)] = C[v7] ⊦ C1[v3 _ v5 _ (v4 _ v5)] _ v0 = v

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ context(O)], [CSIMPSEQ : ⊦ t _ t0 = t1 _ t2 ↔ t = t1 ∧ t0 = t2])]]