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Proof step 10 from proof btree-clean-fwd-strong (previous step)

((btree(rp0, r0, adr, height +1, false, ri0, iast0, alladrs0, alladrs0, nstat) * P0))(zh0), ((btree(rp1, r1, adr, height1, false, ri1, iast1, alladrs1, alladrs1, nstat) * P1))(zh1), iast0 ⊆ iast1, alladrs0 ⊆ dom iast0, alladrs1 ⊆ dom iast1, ∀ rp1, r1, adr, height1, ri1, iast1, alladrs1, nstat, P1, zh1, rp0, r0, ri0, iast0, alladrs0, P0, zh0. ((btree(rp1, r1, adr, height1, false, ri1, iast1, alladrs1, alladrs1, nstat) * P1))(zh1) ∧ ((btree(rp0, r0, adr, height, false, ri0, iast0, alladrs0, alladrs0, nstat) * P0))(zh0) ∧ alladrs1 ⊆ dom iast1 ∧ alladrs0 ⊆ dom iast0 ∧ iast0 ⊆ iast1 → height = height1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1 ⊦ height +1 = height1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
((btree(rp1, r1, adr, height1, false, ri1, iast1, alladrs1, alladrs1, nstat) * P1))(zh1), ((btree(rp0, r0, adr, height + 1, false, ri0, iast0, alladrs0, alladrs0, nstat) * P0))(zh0), alladrs1 ⊆ dom iast1, alladrs0 ⊆ dom iast0, iast0 ⊆ iast1, ∀ rp1, r1, adr, height1, ri1, iast1, alladrs1, nstat, P1, zh1, rp0, r0, ri0, iast0, alladrs0, P0, zh0. ((btree(rp1, r1, adr, height1, false, ri1, iast1, alladrs1, alladrs1, nstat) * P1))(zh1) ∧ ((btree(rp0, r0, adr, height, false, ri0, iast0, alladrs0, alladrs0, nstat) * P0))(zh0) ∧ alladrs1 ⊆ dom iast1 ∧ alladrs0 ⊆ dom iast0 ∧ iast0 ⊆ iast1 → height = height1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1 ⊦ height + 1 = height1 ∧ ri0 = ri1 ∧ alladrs0 = alladrs1

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)