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Proof step 46 from proof subst-SK-equal (previous step)

∀ v3, v4, v5, C0, v, v0. C1[S _ v3 _ v4 _ v5] = C0[K _ v _ v0] ∧ value(v3) ∧ value(v4) ∧ value(v5) ∧ value(v) ∧ value(v0) ∧ context(C0) ∧ context(C1) → C1[v3 _ v5 _ (v4 _ v5)] = C0[v], (t _ C1)[S _ v3 _ v4 _ v5] = C0[K _ v _ v0], value(v3), value(v4), value(v5), value(v), value(v0), context(C0), context(t _ C1) ⊦ (t _ C1)[v3 _ v5 _ (v4 _ v5)] = C0[v]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
⊦ (t1 _ C)[t] = t1 _ C[t]
Proof step 45:
t _ C1[S _ v3 _ v4 _ v5] = C0[K _ v _ v0], value(v3), value(v4), value(v5), value(v), value(v0), context(C0), context(C1), value(t), ∀ 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] ⊦ t _ C1[v3 _ v5 _ (v4 _ v5)] = C0[v]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ context(t _ C) ↔ ¬ ¬ (context(C) ∧ value(t))])]]