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Proof step 16 from proof subst-SS-equal (previous step)

C0 = C _ t, S _ v4 _ v5 _ v6 = C0[S _ v1 _ v2 _ v3], value(v4), value(v5), value(v6), value(v1), value(v2), value(v3), context(C0) ⊦ v4 _ v6 _ (v5 _ v6) = C0[v1 _ v3 _ (v2 _ v3)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ (C _ t1)[t] = C[t] _ t1
Proof step 15:
S _ v4 _ v5 = C[S _ v1 _ v2 _ v3], context(C), value(v3), value(v2), value(v1), value(t), value(v5), value(v4) ⊦ v4 _ t _ (v5 _ t) = C[v1 _ v3 _ (v2 _ v3)] _ t

Rule argument

[emptyarg]

Additional information

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