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Proof step 40 from proof subst-KK-equal (previous step)

∀ v1, v2, C0, v, v0. C1[K _ v1 _ v2] = C0[K _ v _ v0] ∧ value(v1) ∧ value(v2) ∧ value(v) ∧ value(v0) ∧ context(C0) ∧ context(C1) → C1[v1] = C0[v], (t _ C1)[K _ v1 _ v2] = C0[K _ v _ v0], value(v1), value(v2), value(v), value(v0), context(C0), context(t _ C1) ⊦ (t _ C1)[v1] = C0[v]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
⊦ (t1 _ C)[t] = t1 _ C[t]
Proof step 39:
t _ C1[K _ v1 _ v2] = C0[K _ v _ v0], value(v1), value(v2), value(v), value(v0), context(C0), context(C1), value(t), ∀ v3, v4, C, v5, v6. C1[K _ v3 _ v4] = C[K _ v5 _ v6] ∧ value(v3) ∧ value(v4) ∧ value(v5) ∧ value(v6) ∧ context(C) → C1[v3] = C[v5] ⊦ t _ C1[v1] = C0[v]

Rule argument

[emptyarg]

Additional information

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