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Proof step 11 from proof in-list-put (previous step)

∀ r, st, r0, sk, val. sks0 = [] ∨ (r -- sks0 ∈ st[r0 -- sk, val] ↔ r -- sks0 ∈ st) ∨ sk ∈ sks0 ⊦ sk0 + sks0 = [], r -- (sk0 + sks0) ∈ st[r0 -- sk, val] ↔ r -- (sk0 + sks0) ∈ st, sk ∈ sk0 + sks0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ r -- sk ' ∈ st ↔ r -- sk ∈ st
Proof step 10:
∀ r, st, r0, sk, val. (r -- sks0 ∈ st[r0 -- sk, val] ↔ r -- sks0 ∈ st) ∨ sk ∈ sks0 ⊦ r -- (sk0 ' + sks0) ∈ st[r0 -- sk, val] ↔ r -- (sk0 ' + sks0) ∈ st, sk = sk0, sks0 = [], sk ∈ sks0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st], [CSIMPSEQ : ⊦ sks + [] = sks], [CSIMPSEQ : ⊦ ¬ sk ∈ sk0 ' + sks ↔ sk ≠ sk0 ∧ ¬ sk ∈ sks], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk ' + sks ≠ []], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]