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Proof step 6 from proof onlymod-put (previous step)

r -- sk ∈ st, r ∈ st, ∀ rk. rk ∈ st ↔ rk ∈ st0, ∀ 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] ⊦ sk = _length, sk = _type, (rk ∈ st → rk = r -- sk ∨ rk ∈ st0) ∧ (r0 ≠ r → st[r0 -- sk0] = st0[r0 -- sk0])

User applied rule all left

And obtained the following new goals

Proof step 5:
r -- sk ∈ st, r ∈ st, rk ∈ st ↔ rk ∈ st0, ∀ rk. rk ∈ st ↔ rk ∈ st0, ∀ 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] ⊦ sk = _length, sk = _type, (rk ∈ st → rk = r -- sk ∨ rk ∈ st0) ∧ (r0 ≠ r → st[r0 -- sk0] = st0[r0 -- sk0])

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (rk)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk)]]