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Proof step 7 from proof onlymod-del (previous step)

rk0 ∈ st ↔ rk0 ∈ st0, rk ∈ st, ∀ rk. rk ∈ st ↔ rk ∈ st0, r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk], ∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk], r -- _type ∈ st → st[r -- _type] = st0[r -- _type], r -- _length ∈ st → st[r -- _length] = st0[r -- _length] ⊦ r0 ≠ r → (st -- rk)[r0 -- sk] = (st0 -- rk)[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
rk0 ∈ st ↔ rk0 ∈ st0, st[r0 -- sk] = st0[r0 -- sk], rk ∈ st, ∀ r0, sk. r0 ≠ r → st[r0 -- sk] = st0[r0 -- sk], r -- _type ∈ st → st[r -- _type] = st0[r -- _type], r -- _length ∈ st → st[r -- _length] = st0[r -- _length], ∀ rk. rk ∈ st ↔ rk ∈ st0 ⊦ (st -- rk)[r0 -- sk] = (st0 -- rk)[r0 -- sk], r0 = r

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]