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Proof step 26 from proof subst-KK-equal (previous step)

C0 = C2 _ t0, C1[K _ v1 _ v2] _ t = C0[K _ v _ v0], value(v1), value(v2), value(v), value(v0), context(C0), context(C1), ∀ 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] ⊦ C1[v1] _ t = C0[v]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
⊦ (C _ t1)[t] = C[t] _ t1
Proof step 25:
C1[K _ v1 _ v2] = C2[K _ v _ v0], context(C2), value(v0), value(v), value(v2), value(v1), context(C1), ∀ 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] ⊦ C1[v1] = C2[v]

Rule argument

[emptyarg]

Additional information

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